Operaciones combinadas de fracciones.

Colección de 44 operaciones combinadas de fracciones.

¿Qué es una operación combinada con fracciones? Es un grupo de fracciones en las que aparecen sumas, restas, multiplicaciones, divisiones, potencias y raíces de fracciones. Veamos unos ejemplos:

$${\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{6}{5}}\text{ y }{\displaystyle \frac{{\displaystyle \frac{5}{3}}-[{\displaystyle \frac{2}{3}}:{\displaystyle \frac{2}{5}}-(3+{\displaystyle \frac{1}{2}})]\cdot {\displaystyle \frac{3}{11}}}{{\displaystyle \frac{14}{3}}-{\displaystyle \frac{13}{3}}:({\displaystyle \frac{2}{5}}-3)+{\displaystyle \frac{1}{2}}}}$$

• Suma y resta de fracciones
  Lo $1^{\underset{―}{o}}$ que nos preguntamos: ¿tienen el mismo denominador?
  Si lo tienen, para sumar o restar fracciones con el mismo denominador, entonces se suman los numeradores como números enteros que son y se deja el mismo denominador.
  Si no lo tienen, reducimos las fracciones a común denominador y se suman o restan las fracciones obtenidas ya con el mismo denominador.
  Hay que tener «mucho cuidado» a la hora de calcular el común denminador usando el mínimo común múltiplo.
  Ejemplos: $${\displaystyle \frac{6}{11}}-{\displaystyle \frac{2}{11}}={\displaystyle \frac{4}{11}}{\displaystyle \frac{4}{13}}+{\displaystyle \frac{3}{13}}={\displaystyle \frac{7}{13}}$$ $${\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}={\displaystyle \frac{3}{6}}-{\displaystyle \frac{2}{6}}={\displaystyle \frac{1}{6}}{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}={\displaystyle \frac{3}{6}}+{\displaystyle \frac{2}{6}}={\displaystyle \frac{5}{6}}$$
• Multiplicar fracciones
  
  
  No tienen que tener el mismo denominador La multiplicación de dos o más fracciones se realiza «en línea», es decir, el numerador del producto será el producto de los numeradores, y el denominador del producto será el producto de los denominadores: Ejemplos: $${\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{7}{4}}={\displaystyle \frac{3\cdot 7}{2\cdot 4}}={\displaystyle \frac{21}{8}}$$ $${\displaystyle \frac{5}{11}}\cdot {\displaystyle \frac{12}{11}}={\displaystyle \frac{5\cdot 12}{11\cdot 11}}={\displaystyle \frac{60}{121}}$$
• Dividir fracciones
  No tienen que tener el mismo denominador
  
  La división de dos fracciones se realiza «en cruz» o «caramelo», es decir, el numerador del cociente será el producto del numerador de la $1^{\underset{―}{a}}$ fracción por el denominador de la $2^{\underset{―}{a}}$ fracción, y el denominador del cociente será el producto del denominador de la $1^{\underset{―}{a}}$ fracción por el numerador de la $2^{\underset{―}{a}}$ fracción: Ejemplos: $${\displaystyle \frac{3}{11}}:{\displaystyle \frac{5}{4}}={\displaystyle \frac{3\cdot 4}{11\cdot 5}}={\displaystyle \frac{12}{55}}$$ $${\displaystyle \frac{4}{5}}:{\displaystyle \frac{7}{9}}={\displaystyle \frac{4\cdot 9}{5\cdot 7}}={\displaystyle \frac{36}{35}}$$

¿Qué pasos seguir para calcular el resultado de una operación combinada? Para saber en que orden debemos realizar las operaciones debemos tener en cuenta la jerarquía de las operaciones:

$$\begin{array}{|cc|}\hline 1^{\underset{―}{o}}& Par\acute{e}ntesis(),corchetes[]ollaves\{\}\\ 2^{\underset{―}{o}}& Potenciasyraíces\\ 3^{\underset{―}{o}}& Multiplicacionesydivisiones\\ 4^{\underset{―}{o}}& Sumasyrestas\\ \hline\end{array}$$

En este caso $1^{\underset{―}{o}}$ haremos el producto y después la suma: $${\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{6}{5}}={\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{\cancel{3}}}\cdot {\displaystyle \frac{{\cancel{6}}^2}{5}}={\displaystyle \frac{1}{4}}+{\displaystyle \frac{2}{5}}={\displaystyle \frac{5}{20}}+{\displaystyle \frac{8}{20}}={\displaystyle \frac{13}{20}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis, es decir, la suma y después el producto: $$({\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{3}})\cdot {\displaystyle \frac{6}{5}}=({\displaystyle \frac{3}{12}}+{\displaystyle \frac{4}{12}})\cdot {\displaystyle \frac{6}{5}}={\displaystyle \frac{7}{12}}\cdot {\displaystyle \frac{6}{5}}={\displaystyle \frac{7}{{\cancel{12}}^2}}\cdot {\displaystyle \frac{\cancel{6}}{5}}={\displaystyle \frac{7}{10}}$$

En este caso $1^{\underset{―}{o}}$ haremos el producto y después la resta: $$1-{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{1}{5}}=1-{\displaystyle \frac{2}{15}}={\displaystyle \frac{15}{15}}-{\displaystyle \frac{2}{15}}={\displaystyle \frac{13}{15}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis, la resta, y después el producto: $$(1-{\displaystyle \frac{2}{3}})\cdot {\displaystyle \frac{1}{5}}=({\displaystyle \frac{3}{3}}-{\displaystyle \frac{2}{3}})\cdot {\displaystyle \frac{1}{5}}={\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{1}{5}}={\displaystyle \frac{1}{15}}$$

En este caso $1^{\underset{―}{o}}$ haremos el producto y después la suma: $$-{\displaystyle \frac{2}{3}}+{\displaystyle \frac{4}{3}}\cdot {\displaystyle \frac{1}{2}}=-{\displaystyle \frac{2}{3}}+{\displaystyle \frac{{\cancel{4}}^2}{3}}\cdot {\displaystyle \frac{1}{\cancel{2}}}=-{\displaystyle \frac{2}{3}}+{\displaystyle \frac{2}{3}}=0$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis y después el producto. La operación del paréntesis se puede hacer poniendo el mismo denominador a todas las fracciones u operación a operación que es como hemos elegido en este caso: $$(-1+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}})\cdot {\displaystyle \frac{6}{5}}=(-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}})\cdot {\displaystyle \frac{6}{5}}=({\displaystyle \frac{-3}{6}}-{\displaystyle \frac{2}{6}})\cdot {\displaystyle \frac{6}{5}}={\displaystyle \frac{-5}{6}}\cdot {\displaystyle \frac{6}{5}}={\displaystyle \frac{-5}{\xcancel{6}}}\cdot {\displaystyle \frac{\xcancel{6}}{5}}={\displaystyle \frac{-5}{5}}=-1$$

En este caso $1^{\underset{―}{o}}$ haremos los productos y después la suma y la resta poniendo a todas las fracciones el mismo denominador: $$-{\displaystyle \frac{2}{5}}+{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{6}{5}}=-{\displaystyle \frac{2}{5}}+{\displaystyle \frac{4}{15}}-{\displaystyle \frac{6}{15}}=-{\displaystyle \frac{6}{15}}+{\displaystyle \frac{4}{15}}-{\displaystyle \frac{6}{15}}={\displaystyle \frac{-12+4}{15}}={\displaystyle \frac{-8}{15}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis, después el producto la suma y por último la resta: $$(-{\displaystyle \frac{2}{5}}+{\displaystyle \frac{1}{3}})\cdot {\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{6}{5}}=(-{\displaystyle \frac{6}{15}}+{\displaystyle \frac{5}{15}})\cdot {\displaystyle \frac{4}{5}}-{\displaystyle \frac{6}{15}}=-{\displaystyle \frac{1}{15}}\cdot {\displaystyle \frac{4}{5}}-{\displaystyle \frac{6}{15}}={\displaystyle \frac{-4}{75}}-{\displaystyle \frac{30}{75}}={\displaystyle \frac{-34}{75}}$$

En este caso $1^{\underset{―}{o}}$ haremos primeros los productos y después poniendo el mismo denominador las sumas y las restas: $${\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{4}{3}}-{\displaystyle \frac{1}{12}}+{\displaystyle \frac{5}{4}}\cdot {\displaystyle \frac{8}{3}}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{4}{9}}-{\displaystyle \frac{1}{12}}+{\displaystyle \frac{5}{\cancel{4}}}\cdot {\displaystyle \frac{{\cancel{8}}^2}{3}}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{4}{9}}-{\displaystyle \frac{1}{12}}+{\displaystyle \frac{10}{3}}=$$
$$={\displaystyle \frac{18}{36}}+{\displaystyle \frac{16}{36}}-{\displaystyle \frac{3}{36}}+{\displaystyle \frac{120}{36}}={\displaystyle \frac{18+16-3+120}{36}}={\displaystyle \frac{151}{36}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis, después el producto que está más a la derecha, después el producto que está a la izquierda y después el resto de operaciones: $$({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}})\cdot {\displaystyle \frac{4}{3}}-{\displaystyle \frac{1}{12}}+{\displaystyle \frac{5}{\cancel{4}}}\cdot {\displaystyle \frac{{\cancel{8}}^2}{3}}=({\displaystyle \frac{3}{6}}+{\displaystyle \frac{2}{6}})\cdot {\displaystyle \frac{4}{3}}-{\displaystyle \frac{1}{12}}+{\displaystyle \frac{10}{3}}={\displaystyle \frac{5}{6}}\cdot {\displaystyle \frac{4}{3}}-{\displaystyle \frac{1}{12}}+{\displaystyle \frac{10}{3}}=$$
$$={\displaystyle \frac{5}{{\cancel{6}}^3}}\cdot {\displaystyle \frac{{\cancel{4}}^2}{3}}-{\displaystyle \frac{1}{12}}+{\displaystyle \frac{10}{3}}={\displaystyle \frac{10}{9}}-{\displaystyle \frac{1}{12}}+{\displaystyle \frac{10}{3}}={\displaystyle \frac{40}{36}}-{\displaystyle \frac{3}{36}}+{\displaystyle \frac{120}{36}}={\displaystyle \frac{157}{36}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis y después el producto. La operación del paréntesis se puede hacer poniendo el mismo denominador a todas las fracciones u operación a operación que es como hemos elegido en este caso: $$(1-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}})\cdot {\displaystyle \frac{2}{5}}=({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}})\cdot {\displaystyle \frac{2}{5}}=({\displaystyle \frac{3}{6}}+{\displaystyle \frac{2}{6}})\cdot {\displaystyle \frac{2}{5}}={\displaystyle \frac{5}{6}}\cdot {\displaystyle \frac{2}{5}}={\displaystyle \frac{\cancel{5}}{6}}\cdot {\displaystyle \frac{2}{\cancel{5}}}={\displaystyle \frac{2}{6}}={\displaystyle \frac{1}{3}}$$

En este caso $1^{\underset{―}{o}}$ haremos la resta, ya qué es muy sencilla, después el producto y para terminar la suma: $$1-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{2}{5}}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{2}{15}}={\displaystyle \frac{15}{30}}+{\displaystyle \frac{4}{30}}={\displaystyle \frac{19}{30}}$$

En este caso $1^{\underset{―}{o}}$ haremos los productos, como queda el mismo denominador haremos la resta y a la suma a la vez. En esta ocasión de nada sirve simplificar: $$-{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{4}{7}}-{\displaystyle \frac{2}{14}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{5}{7}}=-{\displaystyle \frac{4}{14}}-{\displaystyle \frac{2}{14}}+{\displaystyle \frac{5}{14}}=-{\displaystyle \frac{1}{14}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis y después los productos y después la suma. El producto que está más a la derecha se puede hacer a la vez que el paréntesis: $$-{\displaystyle \frac{1}{2}}\cdot ({\displaystyle \frac{4}{7}}-{\displaystyle \frac{2}{14}})+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{5}{7}}=-{\displaystyle \frac{1}{2}}\cdot ({\displaystyle \frac{8}{14}}-{\displaystyle \frac{2}{14}})+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{5}{7}}=$$
$$=-{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{6}{14}}+{\displaystyle \frac{5}{14}}=-{\displaystyle \frac{1}{\cancel{2}}}\cdot {\displaystyle \frac{{\cancel{6}}^3}{14}}+{\displaystyle \frac{5}{14}}=-{\displaystyle \frac{3}{14}}+{\displaystyle \frac{5}{14}}={\displaystyle \frac{2}{14}}={\displaystyle \frac{1}{7}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis y después simplificaremos las fracciones que podamos, luego la división y por último el resto de operaciones: $${\displaystyle \frac{17}{9}}-{\displaystyle \frac{15}{5}}+{\displaystyle \frac{4}{3}}:({\displaystyle \frac{1}{5}}+{\displaystyle \frac{2}{3}}-{\displaystyle \frac{1}{15}})+{\displaystyle \frac{14}{3}}:{\displaystyle \frac{16}{8}}={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{4}{3}}:({\displaystyle \frac{3}{15}}+{\displaystyle \frac{10}{15}}-{\displaystyle \frac{1}{15}})+{\displaystyle \frac{14}{3}}:2=$$
$$={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{4}{3}}:{\displaystyle \frac{12}{15}}+{\displaystyle \frac{14}{6}}={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{4}{3}}:{\displaystyle \frac{{\cancel{12}}^4}{{\cancel{15}}^5}}+{\displaystyle \frac{14}{6}}={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{4}{3}}:{\displaystyle \frac{4}{5}}+{\displaystyle \frac{7}{3}}=$$
$$={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{5}{3}}+{\displaystyle \frac{7}{3}}={\displaystyle \frac{17}{9}}-3+4={\displaystyle \frac{17}{9}}+1={\displaystyle \frac{17}{9}}+{\displaystyle \frac{9}{9}}={\displaystyle \frac{26}{9}}$$

En este caso $1^{\underset{―}{o}}$ haremos a la vez la división y el paréntesis, simplificaremos todo lo que podamos. Después haremos el producto y al final la suma: $${\displaystyle \frac{1}{3}}+{\displaystyle \frac{4}{3}}:{\displaystyle \frac{5}{6}}\cdot ({\displaystyle \frac{1}{2}}-{\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{10}{9}}+4)={\displaystyle \frac{1}{3}}+{\displaystyle \frac{4\cdot {\cancel{6}}^2}{\cancel{3}\cdot 5}}\cdot ({\displaystyle \frac{1}{2}}-{\displaystyle \frac{\cancel{3}}{2}}\cdot {\displaystyle \frac{10}{{\cancel{9}}^3}}+4)={\displaystyle \frac{1}{3}}+{\displaystyle \frac{8}{5}}\cdot ({\displaystyle \frac{1}{2}}-{\displaystyle \frac{\cancel{10}5}{\cancel{2}\cdot 3}}+4)=$$
$$={\displaystyle \frac{1}{3}}+{\displaystyle \frac{8}{5}}\cdot ({\displaystyle \frac{1}{2}}-{\displaystyle \frac{5}{3}}+4)={\displaystyle \frac{1}{3}}+{\displaystyle \frac{8}{5}}\cdot ({\displaystyle \frac{3}{6}}-{\displaystyle \frac{10}{6}}+{\displaystyle \frac{24}{6}})={\displaystyle \frac{1}{3}}+{\displaystyle \frac{8}{5}}\cdot {\displaystyle \frac{17}{6}}=$$
$$={\displaystyle \frac{1}{3}}+{\displaystyle \frac{\cancel{8}4}{5}}\cdot {\displaystyle \frac{17}{{\cancel{6}}^3}}={\displaystyle \frac{1}{3}}+{\displaystyle \frac{4\cdot 17}{5\cdot 3}}={\displaystyle \frac{5}{15}}+{\displaystyle \frac{68}{15}}={\displaystyle \frac{73}{15}}$$

En este caso $1^{\underset{―}{o}}$ haremos a la vez el producto, el paréntesis y la divisón, simplificaremos todo lo que podamos. Después haremos el producto y al final el resto de operaciones: $${\displaystyle \frac{4}{5}}-{\displaystyle \frac{7}{3}}\cdot {\displaystyle \frac{3}{7}}+{\displaystyle \frac{1}{5}}\cdot (2+{\displaystyle \frac{1}{2}})-{\displaystyle \frac{7}{3}}+4:{\displaystyle \frac{6}{5}}={\displaystyle \frac{4}{5}}-{\displaystyle \frac{\xcancel{7}}{\cancel{3}}}\cdot {\displaystyle \frac{\cancel{3}}{\xcancel{7}}}+{\displaystyle \frac{1}{5}}\cdot ({\displaystyle \frac{4}{2}}+{\displaystyle \frac{1}{2}})-{\displaystyle \frac{7}{3}}+{\displaystyle \frac{4\cdot 5}{6}}=$$
$$={\displaystyle \frac{4}{5}}-1+{\displaystyle \frac{1}{5}}\cdot {\displaystyle \frac{5}{2}}-{\displaystyle \frac{7}{3}}+{\displaystyle \frac{{\cancel{4}}^2\cdot 5}{{\cancel{6}}^3}}={\displaystyle \frac{4}{5}}-1+{\displaystyle \frac{1}{\cancel{5}}}\cdot {\displaystyle \frac{\cancel{5}}{2}}-{\displaystyle \frac{7}{3}}+{\displaystyle \frac{10}{3}}={\displaystyle \frac{4}{5}}-1+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{7}{3}}+{\displaystyle \frac{10}{3}}=$$
$$={\displaystyle \frac{4}{5}}-1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{3}{3}}={\displaystyle \frac{4}{5}}-1+{\displaystyle \frac{1}{2}}+1={\displaystyle \frac{4}{5}}+{\displaystyle \frac{1}{2}}={\displaystyle \frac{4}{5}}\xcancel{-1}+{\displaystyle \frac{1}{2}}\xcancel{+1}={\displaystyle \frac{4}{5}}+{\displaystyle \frac{1}{2}}={\displaystyle \frac{8}{10}}+{\displaystyle \frac{5}{10}}={\displaystyle \frac{13}{10}}$$

En este caso $1^{\underset{―}{o}}$ haremos los paréntesis, después el producto y por último el resto de operaciones: $${\displaystyle \frac{2}{3}}+{\displaystyle \frac{5}{4}}\cdot ({\displaystyle \frac{3}{5}}+{\displaystyle \frac{4}{10}})-{\displaystyle \frac{5}{4}}+({\displaystyle \frac{3}{5}}:4)+{\displaystyle \frac{12}{5}}={\displaystyle \frac{2}{3}}+{\displaystyle \frac{5}{4}}\cdot ({\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}})-{\displaystyle \frac{5}{4}}+{\displaystyle \frac{3}{20}}+{\displaystyle \frac{12}{5}}={\displaystyle \frac{2}{3}}+{\displaystyle \frac{5}{4}}\cdot 1-{\displaystyle \frac{5}{4}}+{\displaystyle \frac{3}{20}}+{\displaystyle \frac{12}{5}}=$$
$$={\displaystyle \frac{2}{3}}+\xcancel{{\displaystyle \frac{5}{4}}}-\xcancel{{\displaystyle \frac{5}{4}}}+{\displaystyle \frac{3}{20}}+{\displaystyle \frac{12}{5}}={\displaystyle \frac{2}{3}}+{\displaystyle \frac{3}{20}}+{\displaystyle \frac{48}{20}}={\displaystyle \frac{2}{3}}+{\displaystyle \frac{51}{20}}={\displaystyle \frac{40}{60}}+{\displaystyle \frac{153}{60}}={\displaystyle \frac{193}{60}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis, después la divisón y por último la suma: $$2+{\displaystyle \frac{1}{5}}:(2+{\displaystyle \frac{7}{3}}-{\displaystyle \frac{2}{4}}+{\displaystyle \frac{5}{3}})=2+{\displaystyle \frac{1}{5}}:({\displaystyle \frac{4}{2}}+{\displaystyle \frac{7}{3}}-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{5}{3}})=2+{\displaystyle \frac{1}{5}}:({\displaystyle \frac{4-1}{2}}+{\displaystyle \frac{7+5}{3}})=$$
$$=2+{\displaystyle \frac{1}{5}}:({\displaystyle \frac{3}{2}}+{\displaystyle \frac{12}{3}})=2+{\displaystyle \frac{1}{5}}:({\displaystyle \frac{3}{2}}+4)=2+{\displaystyle \frac{1}{5}}:\left({\displaystyle \frac{3+8}{2}}\right)=2+{\displaystyle \frac{1}{5}}:{\displaystyle \frac{11}{2}}=2+{\displaystyle \frac{2}{55}}={\displaystyle \frac{110}{55}}+{\displaystyle \frac{2}{55}}={\displaystyle \frac{112}{55}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis y la divisón, después el producto y por último la suma: $$({\displaystyle \frac{2}{7}}-{\displaystyle \frac{4}{5}}+{\displaystyle \frac{2}{8}})\cdot {\displaystyle \frac{3}{2}}-{\displaystyle \frac{7}{5}}:{\displaystyle \frac{4}{7}}=({\displaystyle \frac{2}{7}}-{\displaystyle \frac{4}{5}}+{\displaystyle \frac{1}{4}})\cdot {\displaystyle \frac{3}{2}}-{\displaystyle \frac{49}{20}}=({\displaystyle \frac{40}{140}}-{\displaystyle \frac{122}{140}}+{\displaystyle \frac{35}{140}})\cdot {\displaystyle \frac{3}{2}}-{\displaystyle \frac{49}{20}}=$$
$$={\displaystyle \frac{-37}{140}}\cdot {\displaystyle \frac{3}{2}}-{\displaystyle \frac{49}{20}}={\displaystyle \frac{-111}{280}}-{\displaystyle \frac{49}{20}}={\displaystyle \frac{-111}{280}}-{\displaystyle \frac{686}{280}}=-{\displaystyle \frac{797}{280}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis y la divisón que está más a la derecha, simplificando siempre que sea posible, después la división del paréntesis y por último las sumas y restas: $${\displaystyle \frac{17}{9}}-{\displaystyle \frac{15}{5}}+{\displaystyle \frac{4}{3}}:({\displaystyle \frac{1}{5}}+{\displaystyle \frac{2}{3}}-{\displaystyle \frac{1}{15}})+{\displaystyle \frac{14}{3}}:{\displaystyle \frac{16}{8}}={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{4}{3}}:({\displaystyle \frac{3}{15}}+{\displaystyle \frac{10}{15}}-{\displaystyle \frac{1}{15}})+{\displaystyle \frac{14}{3}}:2$$
$$={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{4}{3}}:{\displaystyle \frac{12}{15}}+{\displaystyle \frac{14}{2\cdot 3}}={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{4}{3}}:{\displaystyle \frac{{\cancel{12}}^4}{{\cancel{15}}^5}}+{\displaystyle \frac{{\cancel{14}}^7}{\cancel{2}\cdot 3}}={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{4}{3}}:{\displaystyle \frac{4}{5}}+{\displaystyle \frac{7}{3}}={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{4\cdot 5}{4\cdot 3}}+{\displaystyle \frac{7}{3}}=$$
$$={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{\cancel{4}\cdot 5}{\cancel{4}\cdot 3}}+{\displaystyle \frac{7}{3}}={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{5}{3}}+{\displaystyle \frac{7}{3}}={\displaystyle \frac{17}{9}}-3+{\displaystyle \frac{12}{3}}={\displaystyle \frac{17}{9}}-3+4={\displaystyle \frac{17}{9}}+1={\displaystyle \frac{17}{9}}+{\displaystyle \frac{9}{9}}={\displaystyle \frac{26}{9}}$$

En este caso $1^{\underset{―}{o}}$ haremos el paréntesis, después el corchete y por último la suma: $${\displaystyle \frac{2}{3}}+[1-({\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{6}})]={\displaystyle \frac{2}{3}}+[1-({\displaystyle \frac{9}{12}}-{\displaystyle \frac{2}{12}})]={\displaystyle \frac{2}{3}}+[1-{\displaystyle \frac{7}{12}}]={\displaystyle \frac{2}{3}}+[{\displaystyle \frac{12}{12}}-{\displaystyle \frac{7}{12}}]={\displaystyle \frac{2}{3}}+{\displaystyle \frac{5}{12}}=$$
$$={\displaystyle \frac{8}{12}}+{\displaystyle \frac{5}{12}}={\displaystyle \frac{13}{12}}$$

En este caso $1^{\underset{―}{o}}$ haremos los paréntesis, después el corchete y por último las sumas y restas: $${\displaystyle \frac{2}{3}}-[{\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{5}}-({\displaystyle \frac{2}{5}}-{\displaystyle \frac{1}{3}})+({\displaystyle \frac{6}{5}}-{\displaystyle \frac{1}{2}})]-{\displaystyle \frac{3}{4}}+({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}})=$$
$$={\displaystyle \frac{2}{3}}-[{\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{5}}-({\displaystyle \frac{6}{15}}-{\displaystyle \frac{5}{15}})+({\displaystyle \frac{12}{10}}-{\displaystyle \frac{5}{10}})]-{\displaystyle \frac{3}{4}}+({\displaystyle \frac{3}{6}}-{\displaystyle \frac{2}{6}})={\displaystyle \frac{2}{3}}-[{\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{15}}+{\displaystyle \frac{7}{10}}]-{\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{6}}$$
$$={\displaystyle \frac{2}{3}}-[{\displaystyle \frac{45}{30}}-{\displaystyle \frac{6}{30}}-{\displaystyle \frac{2}{30}}+{\displaystyle \frac{21}{30}}]-{\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{6}}={\displaystyle \frac{2}{3}}-{\displaystyle \frac{58}{30}}-{\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{6}}={\displaystyle \frac{40}{60}}-{\displaystyle \frac{116}{60}}-{\displaystyle \frac{45}{60}}+{\displaystyle \frac{10}{60}}=$$
$$={\displaystyle \frac{-111}{60}}=-{\displaystyle \frac{37}{20}}$$

En este caso $1^{\underset{―}{o}}$ haremos los paréntesis, después el corchete y por último la suma y la resta: $$2+({\displaystyle \frac{5}{2}}-3)-[{\displaystyle \frac{7}{10}}-({\displaystyle \frac{2}{5}}+{\displaystyle \frac{1}{4}})]=2+({\displaystyle \frac{5}{2}}-{\displaystyle \frac{6}{2}})-[{\displaystyle \frac{7}{10}}-({\displaystyle \frac{8}{20}}+{\displaystyle \frac{5}{20}})]=2+(-{\displaystyle \frac{1}{2}})-[{\displaystyle \frac{7}{10}}-{\displaystyle \frac{13}{20}}]=$$
$$2-{\displaystyle \frac{1}{2}}-[{\displaystyle \frac{7}{10}}-{\displaystyle \frac{13}{20}}]={\displaystyle \frac{4}{2}}-{\displaystyle \frac{1}{2}}-[{\displaystyle \frac{14}{20}}-{\displaystyle \frac{13}{20}}]={\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{20}}={\displaystyle \frac{30}{20}}-{\displaystyle \frac{1}{20}}={\displaystyle \frac{29}{20}}$$

En este caso $1^{\underset{―}{o}}$ haremos los paréntesis, después el corchete y por último las restas: $$2-[{\displaystyle \frac{4}{3}}-({\displaystyle \frac{1}{2}}+{\displaystyle \frac{2}{5}})-{\displaystyle \frac{1}{3}}]-({\displaystyle \frac{4}{3}}+2)-{\displaystyle \frac{1}{5}}=2-[{\displaystyle \frac{4}{3}}-({\displaystyle \frac{5}{10}}+{\displaystyle \frac{4}{10}})-{\displaystyle \frac{1}{3}}]-({\displaystyle \frac{4}{3}}+{\displaystyle \frac{6}{3}})-{\displaystyle \frac{1}{5}}=$$

$$=2-[{\displaystyle \frac{4}{3}}-{\displaystyle \frac{9}{10}}-{\displaystyle \frac{1}{3}}]-{\displaystyle \frac{10}{3}}-{\displaystyle \frac{1}{5}}=2-[1-{\displaystyle \frac{9}{10}}]-{\displaystyle \frac{10}{3}}-{\displaystyle \frac{1}{5}}=2-{\displaystyle \frac{1}{10}}-{\displaystyle \frac{10}{3}}-{\displaystyle \frac{1}{5}}={\displaystyle \frac{60}{30}}-{\displaystyle \frac{3}{30}}-{\displaystyle \frac{100}{30}}-{\displaystyle \frac{6}{30}}={\displaystyle \frac{-49}{30}}$$

En este caso $1^{\underset{―}{o}}$ haremos los paréntesis, después el corchete y por último la suma y la resta: $$({\displaystyle \frac{4}{3}}-{\displaystyle \frac{-1}{9}})+[2-({\displaystyle \frac{-5}{4}}+{\displaystyle \frac{2}{3}})]-{\displaystyle \frac{7}{2}}=({\displaystyle \frac{12}{9}}+{\displaystyle \frac{1}{9}})+[2-({\displaystyle \frac{-15}{12}}+{\displaystyle \frac{8}{12}})]-{\displaystyle \frac{7}{2}}={\displaystyle \frac{13}{9}}+[2-\left({\displaystyle \frac{-7}{12}}\right)]-{\displaystyle \frac{7}{2}}=$$
$$={\displaystyle \frac{13}{9}}+[{\displaystyle \frac{24}{12}}+{\displaystyle \frac{7}{12}}]-{\displaystyle \frac{7}{2}}={\displaystyle \frac{13}{9}}+{\displaystyle \frac{31}{12}}-{\displaystyle \frac{7}{2}}={\displaystyle \frac{52}{36}}+{\displaystyle \frac{93}{36}}-{\displaystyle \frac{126}{36}}={\displaystyle \frac{19}{36}}$$

En este caso $1^{\underset{―}{o}}$ haremos los paréntesis, después la divisón del corchete y por último el producto: $$[({\displaystyle \frac{4}{6}}+{\displaystyle \frac{1/7}{2}}):({\displaystyle \frac{4}{3}}-{\displaystyle \frac{5}{12}})]\cdot ({\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{15}})=[({\displaystyle \frac{2}{3}}+{\displaystyle \frac{1}{14}}):({\displaystyle \frac{16}{12}}-{\displaystyle \frac{5}{12}})]\cdot ({\displaystyle \frac{5}{30}}+{\displaystyle \frac{2}{30}})=$$
$$=[({\displaystyle \frac{28}{42}}+{\displaystyle \frac{3}{42}}):{\displaystyle \frac{11}{12}}]\cdot {\displaystyle \frac{7}{30}}={\displaystyle \frac{31}{42}}:{\displaystyle \frac{11}{12}}\cdot {\displaystyle \frac{7}{30}}={\displaystyle \frac{31\cdot 12\cdot 7}{42\cdot 11\cdot 30}}={\displaystyle \frac{31\cdot {\cancel{12}}^2\cdot 7}{{\cancel{42}}^7\cdot 11\cdot 30}}={\displaystyle \frac{31\cdot 2\cdot \xcancel{7}}{\xcancel{7}\cdot 11\cdot 30}}={\displaystyle \frac{31\cdot 2}{11\cdot 30}}=$$
$$={\displaystyle \frac{31\cdot \cancel{2}}{11\cdot {\cancel{30}}^{15}}}={\displaystyle \frac{31}{11\cdot 15}}={\displaystyle \frac{31}{165}}$$

En este caso $1^{\underset{―}{o}}$ haremos los paréntesis, después llos corchetes y por último la resta: $$[-{\displaystyle \frac{3}{8}}+(4-{\displaystyle \frac{1}{2}})]-[(2-{\displaystyle \frac{5}{4}})+({\displaystyle \frac{7}{2}}-{\displaystyle \frac{1}{8}})]=[-{\displaystyle \frac{3}{8}}+({\displaystyle \frac{8}{2}}-{\displaystyle \frac{1}{2}})]-[({\displaystyle \frac{8}{4}}-{\displaystyle \frac{5}{4}})+({\displaystyle \frac{28}{8}}-{\displaystyle \frac{1}{8}})]=$$

$$=[-{\displaystyle \frac{3}{8}}+{\displaystyle \frac{7}{2}}]-[{\displaystyle \frac{3}{4}}+{\displaystyle \frac{27}{8}}]=[-{\displaystyle \frac{3}{8}}+{\displaystyle \frac{28}{8}}]-[{\displaystyle \frac{6}{8}}+{\displaystyle \frac{27}{8}}]={\displaystyle \frac{25}{8}}-{\displaystyle \frac{33}{8}}={\displaystyle \frac{-8}{8}}=-1$$

$$({\displaystyle \frac{1}{3}}-{\displaystyle \frac{4}{5}})\cdot [({\displaystyle \frac{1}{3}}-1)\cdot 3-{\displaystyle \frac{1+2/5}{3}}]=(\ast )$$

Vamos a cambiar la forma de enfocar este tipo de ejercicios, primero hacemos el primer paréntesis, es decir, ${\displaystyle \frac{1}{3}}-{\displaystyle \frac{4}{5}}={\displaystyle \frac{5}{15}}-{\displaystyle \frac{12}{15}}={\displaystyle \frac{-7}{15}}$

continuamos por el corchete, haciendo la operación $({\displaystyle \frac{1}{3}}-1)\cdot 3=({\displaystyle \frac{1}{3}}-{\displaystyle \frac{3}{3}})\cdot 3={\displaystyle \frac{-2\cdot 3}{3}}=-2$

y por último hacemos la operación que queda: ${\displaystyle \frac{1+2/5}{3}}={\displaystyle \frac{5/5+2/5}{3}}={\displaystyle \frac{7/5}{3}}={\displaystyle \frac{7}{15}}$

Ahora continuamos en (*) sustituyendo todas las operaciones por sus resultados:

$$(\ast )={\displaystyle \frac{-7}{15}}\cdot (-2-{\displaystyle \frac{7}{15}})={\displaystyle \frac{-7}{15}}\cdot (-{\displaystyle \frac{30}{15}}-{\displaystyle \frac{7}{15}})={\displaystyle \frac{-7}{15}}\cdot \left({\displaystyle \frac{-37}{15}}\right)={\displaystyle \frac{259}{225}}$$

$${\displaystyle \frac{4}{5}}:[{\displaystyle \frac{12}{16}}\cdot ({\displaystyle \frac{1}{6}}+{\displaystyle \frac{2}{3}})-{\displaystyle \frac{3}{8}}]-3\cdot [{\displaystyle \frac{1}{6}}:(1-{\displaystyle \frac{2}{5}})]={\displaystyle \frac{4}{5}}:[{\displaystyle \frac{3}{4}}\cdot ({\displaystyle \frac{1}{6}}+{\displaystyle \frac{4}{6}})-{\displaystyle \frac{3}{8}}]-3\cdot [{\displaystyle \frac{1}{6}}:({\displaystyle \frac{5}{5}}-{\displaystyle \frac{2}{5}})]=$$

$$={\displaystyle \frac{4}{5}}:[{\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{5}{6}}-{\displaystyle \frac{3}{8}}]-3\cdot [{\displaystyle \frac{1}{6}}:{\displaystyle \frac{3}{5}}]={\displaystyle \frac{4}{5}}:[{\displaystyle \frac{5}{8}}-{\displaystyle \frac{3}{8}}]-3\cdot {\displaystyle \frac{5}{18}}={\displaystyle \frac{4}{5}}:{\displaystyle \frac{2}{8}}-{\displaystyle \frac{5}{6}}={\displaystyle \frac{32}{10}}-{\displaystyle \frac{5}{6}}={\displaystyle \frac{96}{30}}-{\displaystyle \frac{25}{30}}={\displaystyle \frac{71}{30}}$$

$${\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{4}{3}}:({\displaystyle \frac{4}{3}}-{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{15}{8}}+1)={\displaystyle \frac{3}{2}}-{\displaystyle \frac{2}{3}}:({\displaystyle \frac{4}{3}}-{\displaystyle \frac{5}{4}}+1)={\displaystyle \frac{3}{2}}-{\displaystyle \frac{2}{3}}:({\displaystyle \frac{16}{12}}-{\displaystyle \frac{15}{12}}+{\displaystyle \frac{12}{12}})=$$

$$={\displaystyle \frac{3}{2}}-{\displaystyle \frac{2}{3}}:{\displaystyle \frac{13}{12}}={\displaystyle \frac{3}{2}}-{\displaystyle \frac{2\cdot 12}{3\cdot 13}}={\displaystyle \frac{3}{2}}-{\displaystyle \frac{8}{13}}={\displaystyle \frac{39}{26}}-{\displaystyle \frac{16}{26}}={\displaystyle \frac{23}{36}}$$

$$[{\displaystyle \frac{{\displaystyle \frac{5}{3}}}{3-{\displaystyle \frac{1}{2}}}}\cdot ({\displaystyle \frac{6}{4}}-{\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{1}{3}})]\cdot ({\displaystyle \frac{1+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{5}{2}}}}+1)=(\ast )$$

Primero haremos la primera fracción dentro del corchete, es decir, ${\displaystyle \frac{{\displaystyle \frac{5}{3}}}{3-{\displaystyle \frac{1}{2}}}}={\displaystyle \frac{{\displaystyle \frac{5}{3}}}{{\displaystyle \frac{6}{2}}-{\displaystyle \frac{1}{2}}}}={\displaystyle \frac{{\displaystyle \frac{5}{3}}}{{\displaystyle \frac{5}{2}}}}={\displaystyle \frac{2}{3}}$;

Después haremos el paréntesis dentro del corchete ${\displaystyle \frac{6}{4}}-{\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{1}{3}}={\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{2}{2}}=1$

Ahora seguimos con el paréntesis derecho, cogemos el numerador de la fracción $1+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}={\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{3}}={\displaystyle \frac{9}{6}}-{\displaystyle \frac{2}{6}}={\displaystyle \frac{7}{6}}$;

Ahora continuamos con el denominador ${\displaystyle \frac{1}{4}}-{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{5}{2}}={\displaystyle \frac{1}{4}}-{\displaystyle \frac{5}{3}}={\displaystyle \frac{3}{12}}-{\displaystyle \frac{20}{12}}={\displaystyle \frac{-17}{12}}$;

Ahora sustituims en (*) los valores obtenidos:

$$(\ast )=[{\displaystyle \frac{2}{3}}\cdot 1]\cdot ({\displaystyle \frac{{\displaystyle \frac{7}{6}}}{{\displaystyle \frac{-17}{12}}}}+1)={\displaystyle \frac{2}{3}}\cdot ({\displaystyle \frac{-7\cdot 12}{17\cdot 6}}+1)={\displaystyle \frac{2}{3}}\cdot ({\displaystyle \frac{-14}{17}}+1)={\displaystyle \frac{2}{3}}\cdot ({\displaystyle \frac{-14}{17}}+{\displaystyle \frac{17}{17}})={\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{3}{17}}={\displaystyle \frac{2}{17}}$$

Este tipo de ejercicios se resuelven haciendo por una lado el numerador y por otro el denominador. Vamos con el numerador:

$${\displaystyle \frac{3}{5}}+{\displaystyle \frac{1}{2}}={\displaystyle \frac{6}{10}}+{\displaystyle \frac{5}{10}}={\displaystyle \frac{11}{10}}$$
Ahora con el denominador:

$${\displaystyle \frac{2}{3}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{4}{6}}-{\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{6}}$$
Susituimos en (*) y tenemos:

$$(1)={\displaystyle \frac{{\displaystyle \frac{11}{10}}}{{\displaystyle \frac{1}{6}}}}={\displaystyle \frac{11}{10}}:{\displaystyle \frac{1}{6}}={\displaystyle \frac{66}{10}}={\displaystyle \frac{33}{5}}$$

$${\displaystyle \frac{1+{\displaystyle \frac{1+{\displaystyle \frac{1}{2}}}{2}}}{1-{\displaystyle \frac{1-{\displaystyle \frac{1}{3}}}{3}}}}=(\ast )$$

Vamos a empezar por el numerador $1+{\displaystyle \frac{1+{\displaystyle \frac{1}{2}}}{2}}$:

$1^{\underset{―}{o}}$ hacemos la suma: $1+{\displaystyle \frac{1}{2}}={\displaystyle \frac{3}{2}}$, luego lo dividimos por 2: ${\displaystyle \frac{{\displaystyle \frac{3}{2}}}{2}}={\displaystyle \frac{3}{4}}$ y por último le sumamos 1: $1+{\displaystyle \frac{3}{4}}={\displaystyle \frac{7}{4}}$

Ahora vamos con el denominador:

$1^{\underset{―}{o}}$ hacemos la resta: $1-{\displaystyle \frac{1}{3}}={\displaystyle \frac{2}{3}}$, luego lo dividimos por 3: ${\displaystyle \frac{{\displaystyle \frac{2}{3}}}{3}}={\displaystyle \frac{2}{9}}$ y por último se lo restamos a 1: $1-{\displaystyle \frac{2}{9}}={\displaystyle \frac{7}{9}}$

Sustituimos en (*) y tenemos que:

$$(\ast )={\displaystyle \frac{{\displaystyle \frac{7}{4}}}{{\displaystyle \frac{7}{9}}}}={\displaystyle \frac{7\cdot 9}{7\cdot 4}}={\displaystyle \frac{\cancel{7}\cdot 9}{\cancel{7}\cdot 4}}={\displaystyle \frac{9}{4}}$$

$${\displaystyle \frac{({\displaystyle \frac{2}{5}}:3+{\displaystyle \frac{1}{2}})\cdot {\displaystyle \frac{1}{3}}-{\displaystyle \frac{2}{7}}}{{\displaystyle \frac{2}{5}}\cdot 3-({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}})\cdot {\displaystyle \frac{2}{7}}}}=(\ast )$$
Vamos a empezar por el numerador: $$({\displaystyle \frac{2}{5}}:3+{\displaystyle \frac{1}{2}})\cdot {\displaystyle \frac{1}{3}}-{\displaystyle \frac{2}{7}}$$ Empezamos por el paréntesis ${\displaystyle \frac{2}{5}}:3+{\displaystyle \frac{1}{2}}={\displaystyle \frac{2}{15}}+{\displaystyle \frac{1}{2}}={\displaystyle \frac{4}{30}}+{\displaystyle \frac{15}{30}}={\displaystyle \frac{19}{30}}$ Multiplicamos por ${\displaystyle \frac{1}{3}}$ y le restamos ${\displaystyle \frac{2}{7}}$

$${\displaystyle \frac{19}{30}}\cdot {\displaystyle \frac{1}{3}}-{\displaystyle \frac{2}{7}}={\displaystyle \frac{19}{90}}-{\displaystyle \frac{2}{7}}={\displaystyle \frac{133}{630}}-{\displaystyle \frac{180}{630}}={\displaystyle \frac{-47}{630}}$$ Ahora vamos a calcular el denominador:

$${\displaystyle \frac{2}{5}}\cdot 3-({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}})\cdot {\displaystyle \frac{2}{7}}$$ Hacemos $1^{\underset{―}{o}}$ el paréntesis:

${\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}={\displaystyle \frac{3}{6}}+{\displaystyle \frac{2}{6}}={\displaystyle \frac{5}{6}}$ y sustituimos:

${\displaystyle \frac{2}{5}}\cdot 3-{\displaystyle \frac{5}{6}}\cdot {\displaystyle \frac{2}{7}}={\displaystyle \frac{6}{5}}-{\displaystyle \frac{5\cdot 2}{6\cdot 7}}={\displaystyle \frac{6}{5}}-{\displaystyle \frac{5}{21}}={\displaystyle \frac{126}{105}}-{\displaystyle \frac{25}{105}}={\displaystyle \frac{101}{105}}$

Sustituimos en (*) y tenemos $$(\ast )={\displaystyle \frac{{\displaystyle \frac{-47}{630}}}{{\displaystyle \frac{17}{105}}}}={\displaystyle \frac{-47}{630}}:{\displaystyle \frac{17}{105}}={\displaystyle \frac{-47\cdot 105}{101\cdot 630}}={\displaystyle \frac{-47}{101\cdot 6}}={\displaystyle \frac{-47}{606}}$$

$${\displaystyle \frac{{\displaystyle \frac{3}{5}}:(1-{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{9}{4}})+3}{[{\displaystyle \frac{1}{7}}\cdot ({\displaystyle \frac{2}{7}}-{\displaystyle \frac{1}{3}})+{\displaystyle \frac{5}{2}}]:{\displaystyle \frac{1}{2}}}}=(\ast )$$
Vamos a por el numerador del fracción:

$${\displaystyle \frac{3}{5}}:(1-{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{9}{4}})+3=$$ Hacemos primero el paréntesis, producto y resta, una vez hecho el paréntesis hacemos la divsión y la resta:

$$={\displaystyle \frac{3}{5}}:(1-{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{9}{4}})+3={\displaystyle \frac{3}{5}}:(1-{\displaystyle \frac{2\cdot 9}{3\cdot 4}})+3={\displaystyle \frac{3}{5}}:(1-{\displaystyle \frac{\cancel{2}\cdot 9}{3\cdot {\cancel{4}}^2}})+3=$$ $$={\displaystyle \frac{3}{5}}:(1-{\displaystyle \frac{9}{3\cdot 2}})+3={\displaystyle \frac{3}{5}}:(1-{\displaystyle \frac{{\cancel{9}}^3}{\cancel{3}\cdot 2}})+3={\displaystyle \frac{3}{5}}:(1-{\displaystyle \frac{3}{2}})+3={\displaystyle \frac{3}{5}}:\left({\displaystyle \frac{-1}{2}}\right)+3=$$ $$={\displaystyle \frac{-6}{5}}+3={\displaystyle \frac{-6}{5}}+{\displaystyle \frac{15}{5}}={\displaystyle \frac{9}{5}}$$ Ahora calculamos el denominador:

$$[{\displaystyle \frac{1}{7}}\cdot ({\displaystyle \frac{2}{7}}-{\displaystyle \frac{1}{3}})+{\displaystyle \frac{5}{2}}]:{\displaystyle \frac{1}{2}}=$$ Hacemos el paréntesis que hay dentro del corchete, el producto, la suma y por último la división:

$$=[{\displaystyle \frac{1}{7}}\cdot ({\displaystyle \frac{6}{21}}-{\displaystyle \frac{7}{21}})+{\displaystyle \frac{5}{2}}]:{\displaystyle \frac{1}{2}}=[{\displaystyle \frac{1}{7}}\cdot \left({\displaystyle \frac{-1}{21}}\right)+{\displaystyle \frac{5}{2}}]:{\displaystyle \frac{1}{2}}=[{\displaystyle \frac{-2}{294}}\cdot +{\displaystyle \frac{735}{294}}]:{\displaystyle \frac{1}{2}}=$$ $$={\displaystyle \frac{733}{294}}:{\displaystyle \frac{1}{2}}={\displaystyle \frac{733\cdot 2}{294}}={\displaystyle \frac{733\cdot \cancel{2}}{{\cancel{294}}^{147}}}={\displaystyle \frac{733}{147}}$$ Sustituimos en $(\ast )$ y tenemos:

$$(\ast )={\displaystyle \frac{{\displaystyle \frac{9}{5}}}{{\displaystyle \frac{733}{147}}}}={\displaystyle \frac{9\cdot 147}{5\cdot 733}}={\displaystyle \frac{1323}{3665}}$$

$${\displaystyle \frac{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{2}{5}}+{\displaystyle \frac{3}{2}}:{\displaystyle \frac{1}{4}}+5}{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}\cdot ({\displaystyle \frac{2}{5}}+{\displaystyle \frac{3}{2}}:{\displaystyle \frac{1}{4}}+5)}}=(\ast )$$
Vamos a por el numerador del fracción:

$${\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{2}{5}}+{\displaystyle \frac{3}{2}}:{\displaystyle \frac{1}{4}}+5=$$ Hacemos el producto, luego la división y el resto de operaciones de izquierda a derecha:

$$={\displaystyle \frac{1}{2}}-{\displaystyle \frac{2}{15}}+{\displaystyle \frac{12}{2}}+5={\displaystyle \frac{15}{30}}-{\displaystyle \frac{4}{30}}+6+5={\displaystyle \frac{11}{30}}+11={\displaystyle \frac{11}{30}}+{\displaystyle \frac{330}{30}}={\displaystyle \frac{341}{30}}$$ Ahora calculamos el denominador:

$${\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}\cdot ({\displaystyle \frac{2}{5}}+{\displaystyle \frac{3}{2}}:{\displaystyle \frac{1}{4}}+5)=$$ Hacemos el paréntesis, la división y las dos sumas, una vez el paréntesis hacemos el producto y la resta:

$$={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}\cdot ({\displaystyle \frac{2}{5}}+{\displaystyle \frac{12}{2}}+5)={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}\cdot ({\displaystyle \frac{2}{5}}+6+5)={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}\cdot ({\displaystyle \frac{2}{5}}+11)={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}\cdot ({\displaystyle \frac{2}{5}}+{\displaystyle \frac{55}{5}})=$$ $$={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{57}{5}}={\displaystyle \frac{1}{2}}-{\displaystyle \frac{57}{3\cdot 5}}={\displaystyle \frac{1}{2}}-{\displaystyle \frac{{\cancel{57}}^{19}}{\cancel{3}\cdot 5}}={\displaystyle \frac{1}{2}}-{\displaystyle \frac{19}{5}}={\displaystyle \frac{5}{10}}-{\displaystyle \frac{38}{10}}={\displaystyle \frac{-33}{10}}$$ Sustituimos en $(\ast )$ y tenemos:

$$(\ast )={\displaystyle \frac{{\displaystyle \frac{341}{30}}}{{\displaystyle \frac{-33}{10}}}}={\displaystyle \frac{-341\cdot 10}{33\cdot 30}}={\displaystyle \frac{-341\cdot \cancel{10}}{33\cdot {\cancel{30}}^3}}={\displaystyle \frac{-341}{33\cdot 3}}={\displaystyle \frac{-{\cancel{341}}^{31}}{{\cancel{33}}^3\cdot 3}}={\displaystyle \frac{-31}{3\cdot 3}}={\displaystyle \frac{-31}{9}}$$

$${\displaystyle \frac{({\displaystyle \frac{1}{2}}:{\displaystyle \frac{1}{3}}+2)\cdot {\displaystyle \frac{2}{5}}-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{3}}:({\displaystyle \frac{2}{3}}+{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{5}{2}})+{\displaystyle \frac{1}{3}}}}=(\ast )$$
Vamos a por el numerador del fracción:

$$({\displaystyle \frac{1}{2}}:{\displaystyle \frac{1}{3}}+2)\cdot {\displaystyle \frac{2}{5}}-{\displaystyle \frac{1}{2}}=$$ Hacemos primero el paréntesis, la división y después la resta, una vez hecho el paréntesis hacemos el producto y la resta:

$$=({\displaystyle \frac{3}{2}}+2)\cdot {\displaystyle \frac{2}{5}}-{\displaystyle \frac{1}{2}}=({\displaystyle \frac{3}{2}}+{\displaystyle \frac{4}{2}})\cdot {\displaystyle \frac{2}{5}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{7}{2}}\cdot {\displaystyle \frac{2}{5}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{7\cdot 2}{2\cdot 5}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{7\cdot \cancel{2}}{\cancel{2}\cdot 5}}-{\displaystyle \frac{1}{2}}=$$ $$={\displaystyle \frac{7}{5}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{14}{10}}-{\displaystyle \frac{5}{10}}={\displaystyle \frac{9}{10}}$$ Ahora el denominador de la fracción:

$${\displaystyle \frac{1}{3}}:({\displaystyle \frac{2}{3}}+{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{5}{2}})+{\displaystyle \frac{1}{3}}=$$ Hacemos primero el paréntesis, producto y después suma, una vez hecho el paréntesis hacemos la sivisión y la suma:

$$={\displaystyle \frac{1}{3}}:({\displaystyle \frac{2}{3}}+{\displaystyle \frac{5}{6}})+{\displaystyle \frac{1}{3}}={\displaystyle \frac{1}{3}}:({\displaystyle \frac{4}{6}}+{\displaystyle \frac{5}{6}})+{\displaystyle \frac{1}{3}}={\displaystyle \frac{1}{3}}:{\displaystyle \frac{9}{6}}+{\displaystyle \frac{1}{3}}={\displaystyle \frac{1}{3}}:{\displaystyle \frac{3}{2}}+{\displaystyle \frac{1}{3}}=$$ $$={\displaystyle \frac{2}{9}}+{\displaystyle \frac{1}{3}}={\displaystyle \frac{2}{9}}+{\displaystyle \frac{3}{9}}={\displaystyle \frac{5}{9}}$$ Sustituimos en $(\ast )$ y tenemos:

$$(\ast )={\displaystyle \frac{{\displaystyle \frac{9}{10}}}{{\displaystyle \frac{5}{9}}}}={\displaystyle \frac{9\cdot 9}{5\cdot 10}}={\displaystyle \frac{81}{50}}$$

$${\displaystyle \frac{{\displaystyle \frac{2}{5}}-{\displaystyle \frac{6}{3}}+{\displaystyle \frac{2}{3}}}{1-{\displaystyle \frac{2}{5}}-{\displaystyle \frac{6}{4}}}}-{\displaystyle \frac{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{2}{3}}+{\displaystyle \frac{6}{5}}}}=(\ast )$$
Vamos a empezar por el numerador de la primera fracción:

$${\displaystyle \frac{2}{5}}-{\displaystyle \frac{6}{3}}+{\displaystyle \frac{2}{3}}={\displaystyle \frac{2}{5}}-{\displaystyle \frac{4}{3}}={\displaystyle \frac{6}{15}}-{\displaystyle \frac{20}{15}}={\displaystyle \frac{-14}{15}}$$ Ahora el denominador de la primera fracción:

$$1-{\displaystyle \frac{2}{5}}-{\displaystyle \frac{6}{4}}=1-{\displaystyle \frac{2}{5}}-{\displaystyle \frac{3}{2}}={\displaystyle \frac{10}{10}}-{\displaystyle \frac{4}{10}}-{\displaystyle \frac{15}{10}}={\displaystyle \frac{-9}{10}}$$ Ahora a por el numerador de la segunda fracción:

$${\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}={\displaystyle \frac{3}{6}}+{\displaystyle \frac{2}{6}}={\displaystyle \frac{5}{6}}$$ Y ahora el denominador de la segunda fracción:

$${\displaystyle \frac{2}{3}}+{\displaystyle \frac{6}{5}}={\displaystyle \frac{10}{15}}+{\displaystyle \frac{18}{15}}={\displaystyle \frac{28}{15}}$$ Sustituimos en $(\ast )$ y tenemos:

$$(\ast )={\displaystyle \frac{{\displaystyle \frac{-14}{15}}}{{\displaystyle \frac{-9}{10}}}}-{\displaystyle \frac{{\displaystyle \frac{5}{6}}}{{\displaystyle \frac{28}{15}}}}={\displaystyle \frac{14\cdot 10}{15\cdot 9}}-{\displaystyle \frac{5\cdot 15}{6\cdot 28}}={\displaystyle \frac{14\cdot {\cancel{10}}^2}{{\cancel{15}}^3\cdot 9}}-{\displaystyle \frac{5\cdot {\cancel{15}}^5}{{\cancel{6}}^2\cdot 28}}={\displaystyle \frac{14\cdot 2}{3\cdot 9}}-{\displaystyle \frac{5\cdot 5}{2\cdot 28}}=$$ $$={\displaystyle \frac{28}{27}}-{\displaystyle \frac{25}{56}}={\displaystyle \frac{1568}{1512}}-{\displaystyle \frac{675}{1512}}={\displaystyle \frac{893}{1512}}$$

$${\displaystyle \frac{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{4}}}{2+{\displaystyle \frac{5}{2}}-{\displaystyle \frac{1}{6}}}}\cdot {\displaystyle \frac{2}{1-{\displaystyle \frac{3}{2-{\displaystyle \frac{1}{4}}}}}}=(\ast )$$
Vamos a empezar por el numerador de la primera fracción:

$${\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{4}}={\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{3}}={\displaystyle \frac{3}{12}}+{\displaystyle \frac{4}{12}}={\displaystyle \frac{7}{12}}$$ Ahora vamos a calcular el denominador de la primera fracción:

$$2+{\displaystyle \frac{5}{2}}-{\displaystyle \frac{1}{6}}={\displaystyle \frac{12}{6}}+{\displaystyle \frac{15}{6}}-{\displaystyle \frac{1}{6}}={\displaystyle \frac{26}{6}}={\displaystyle \frac{13}{3}}$$ Ahora vamos a calcular el denominador de la segunda fracción:

$$1-{\displaystyle \frac{3}{2-{\displaystyle \frac{1}{4}}}}=1-{\displaystyle \frac{3}{{\displaystyle \frac{8}{4}}-{\displaystyle \frac{1}{4}}}}=1-{\displaystyle \frac{3}{{\displaystyle \frac{7}{4}}}}=1-{\displaystyle \frac{12}{7}}={\displaystyle \frac{7}{7}}-{\displaystyle \frac{12}{7}}={\displaystyle \frac{-5}{7}}$$ Sustituimos en (*) y tenemos:

$$(\ast )={\displaystyle \frac{{\displaystyle \frac{7}{12}}}{{\displaystyle \frac{13}{3}}}}\cdot {\displaystyle \frac{2}{{\displaystyle \frac{-5}{7}}}}={\displaystyle \frac{7\cdot 3}{12\cdot 13}}\cdot {\displaystyle \frac{-14}{5}}={\displaystyle \frac{-14\cdot 7\cdot 3}{12\cdot 13\cdot 5}}={\displaystyle \frac{-14\cdot 7\cdot \cancel{3}}{{\cancel{12}}^4\cdot 13\cdot 5}}={\displaystyle \frac{-14\cdot 7}{4\cdot 13\cdot 5}}=$$ $$={\displaystyle \frac{-{\cancel{14}}^7\cdot 7}{{\cancel{4}}^2\cdot 13\cdot 5}}={\displaystyle \frac{-7\cdot 7}{2\cdot 13\cdot 5}}={\displaystyle \frac{-49}{130}}$$

$${\displaystyle \frac{{\displaystyle \frac{5}{3}}+{\displaystyle \frac{3}{4}}:1-{\displaystyle \frac{5}{4}}+{\displaystyle \frac{17}{3}}}{{\displaystyle \frac{15}{3}}+{\displaystyle \frac{2}{5}}}}=(\ast )$$
Vamos a empezar por el numerador: $${\displaystyle \frac{5}{3}}+{\displaystyle \frac{3}{4}}:1-{\displaystyle \frac{5}{4}}+{\displaystyle \frac{17}{3}}=$$ Empezamos por la división y agrupamos las fracciones por los denominadores de las fracciones: $${\displaystyle \frac{5}{3}}+{\displaystyle \frac{3}{4}}-{\displaystyle \frac{5}{4}}+{\displaystyle \frac{17}{3}}={\displaystyle \frac{5}{3}}-{\displaystyle \frac{2}{4}}+{\displaystyle \frac{17}{3}}={\displaystyle \frac{22}{3}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{44}{6}}-{\displaystyle \frac{3}{6}}={\displaystyle \frac{41}{6}}$$ Ahora vamos a calcular el denominador:

$${\displaystyle \frac{15}{3}}+{\displaystyle \frac{2}{5}}=$$ Hacemos la suma:

$$=5+{\displaystyle \frac{2}{5}}={\displaystyle \frac{25}{5}}+{\displaystyle \frac{2}{5}}={\displaystyle \frac{27}{5}}$$ Sustituimos en (*) y tenemos: $$(\ast )={\displaystyle \frac{{\displaystyle \frac{41}{6}}}{{\displaystyle \frac{27}{5}}}}={\displaystyle \frac{41\cdot 5}{27\cdot 6}}={\displaystyle \frac{205}{162}}$$

$${\displaystyle \frac{[-3+{\displaystyle \frac{2}{5}}\cdot ({\displaystyle \frac{1}{2}}+{\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{8}{27}})]:{\displaystyle \frac{3}{2}}}{({\displaystyle \frac{2}{5}}-3:{\displaystyle \frac{3}{2}})\cdot {\displaystyle \frac{8}{27}}\cdot ({\displaystyle \frac{1}{2}}+{\displaystyle \frac{3}{2}})}}=(\ast )$$
Vamos a empezar por el numerador: $$[-3+{\displaystyle \frac{2}{5}}\cdot ({\displaystyle \frac{1}{2}}+{\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{8}{27}})]:{\displaystyle \frac{3}{2}}=$$ Empezamos por el paréntesis del corchete: $$[-3+{\displaystyle \frac{2}{5}}\cdot ({\displaystyle \frac{1}{2}}+{\displaystyle \frac{\cancel{3}}{2}}\cdot {\displaystyle \frac{8}{{\cancel{27}}^9}})]:{\displaystyle \frac{3}{2}}=[-3+{\displaystyle \frac{2}{5}}\cdot ({\displaystyle \frac{1}{2}}+{\displaystyle \frac{8}{18}})]:{\displaystyle \frac{3}{2}}=[-3+{\displaystyle \frac{2}{5}}\cdot ({\displaystyle \frac{9}{18}}+{\displaystyle \frac{8}{18}})]:{\displaystyle \frac{3}{2}}=$$ $$=[-3+{\displaystyle \frac{2}{5}}\cdot {\displaystyle \frac{17}{18}}]:{\displaystyle \frac{3}{2}}=[-3+{\displaystyle \frac{2\cdot 17}{5\cdot 18}}]:{\displaystyle \frac{3}{2}}=[-3+{\displaystyle \frac{\cancel{2}\cdot 17}{5\cdot {\cancel{18}}^9}}]:{\displaystyle \frac{3}{2}}=[{\displaystyle \frac{-135}{45}}+{\displaystyle \frac{17}{45}}]:{\displaystyle \frac{3}{2}}=$$ $$=\left[{\displaystyle \frac{-118}{45}}\right]:{\displaystyle \frac{3}{2}}={\displaystyle \frac{-118\cdot 2}{45\cdot 3}}={\displaystyle \frac{-236}{135}}$$ Ahora vamos a calcular el denominador:

$$({\displaystyle \frac{2}{5}}-3:{\displaystyle \frac{3}{2}}){\displaystyle \frac{8}{27}}({\displaystyle \frac{1}{2}}+{\displaystyle \frac{3}{2}})=$$ Hacemos los paréntesis y después los productos:

$$=({\displaystyle \frac{2}{5}}-{\displaystyle \frac{2\cdot 3}{3}})\cdot {\displaystyle \frac{8}{27}}\cdot {\displaystyle \frac{4}{2}}=({\displaystyle \frac{2}{5}}-{\displaystyle \frac{2\cdot \cancel{3}}{\cancel{3}}})\cdot {\displaystyle \frac{8}{27}}\cdot 2=({\displaystyle \frac{2}{5}}-2)\cdot {\displaystyle \frac{8}{27}}\cdot 2=({\displaystyle \frac{2}{5}}-{\displaystyle \frac{10}{5}})\cdot {\displaystyle \frac{8}{27}}\cdot 2=$$ $$={\displaystyle \frac{-8}{5}}\cdot {\displaystyle \frac{8}{27}}\cdot 2={\displaystyle \frac{-8\cdot 8\cdot 2}{5\cdot 27}}={\displaystyle \frac{-128}{135}}$$ Sustituimos en (*) y tenemos: $$(\ast )={\displaystyle \frac{{\displaystyle \frac{-236}{135}}}{{\displaystyle \frac{-128}{135}}}}={\displaystyle \frac{-236\cdot 135}{-128\cdot 135}}={\displaystyle \frac{-236\cdot \cancel{135}}{-128\cdot \cancel{135}}}={\displaystyle \frac{236}{128}}={\displaystyle \frac{{\cancel{236}}^{59}}{{\cancel{128}}^{32}}}={\displaystyle \frac{59}{32}}$$

$${\displaystyle \frac{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{2}{4}}+{\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{2}{9}}}{2+{\displaystyle \frac{1}{3}}\cdot (2-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{6}{5}})}}=(\ast )$$
Primero hacemos la operación combinada del numerador:

$${\displaystyle \frac{1}{4}}+{\displaystyle \frac{2}{4}}+{\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{2}{9}}=$$ Podemos hacer la primera suma, antes de hacer el producto simplificamos y terminamos el numerador:

$$={\displaystyle \frac{3}{4}}+{\displaystyle \frac{\cancel{3}}{{\cancel{4}}^2}}\cdot {\displaystyle \frac{\cancel{2}}{{\cancel{9}}^3}}={\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{6}}={\displaystyle \frac{9}{12}}+{\displaystyle \frac{2}{12}}={\displaystyle \frac{11}{12}}$$ Ahora vamos con el denominador:

$$2+{\displaystyle \frac{1}{3}}\cdot (2-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{6}{5}})=$$ Hacemos el paréntesis simplificando lo primero y luego el resto de operaciones:

$$=2+{\displaystyle \frac{1}{3}}\cdot (2-{\displaystyle \frac{1}{\cancel{3}}}\cdot {\displaystyle \frac{{\cancel{6}}^2}{5}})=2+{\displaystyle \frac{1}{3}}\cdot (2-{\displaystyle \frac{2}{5}})=2+{\displaystyle \frac{1}{3}}\cdot ({\displaystyle \frac{10}{5}}-{\displaystyle \frac{2}{5}})=2+{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{8}{5}}=2+{\displaystyle \frac{8}{15}}=$$ $$={\displaystyle \frac{30}{15}}+{\displaystyle \frac{8}{15}}={\displaystyle \frac{38}{15}}$$ Sustituimos en (*) y tenemos

$${\displaystyle \frac{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{2}{4}}+{\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{2}{9}}}{2+{\displaystyle \frac{1}{3}}\cdot (2-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{6}{5}})}}={\displaystyle \frac{{\displaystyle \frac{11}{12}}}{{\displaystyle \frac{38}{15}}}}={\displaystyle \frac{11\cdot 15}{12\cdot 38}}={\displaystyle \frac{11\cdot {\cancel{15}}^5}{{\cancel{12}}^4\cdot 38}}={\displaystyle \frac{11\cdot 5}{4\cdot 38}}={\displaystyle \frac{55}{152}}$$

$${\displaystyle \frac{{\displaystyle \frac{5}{3}}-[{\displaystyle \frac{2}{3}}:{\displaystyle \frac{2}{5}}-(3+{\displaystyle \frac{1}{2}})]\cdot {\displaystyle \frac{3}{11}}}{{\displaystyle \frac{14}{3}}-{\displaystyle \frac{13}{3}}:({\displaystyle \frac{2}{5}}-3)+{\displaystyle \frac{1}{2}}}}=(\ast )$$
Primero hacemos la operación combinada del numerador:

$${\displaystyle \frac{5}{3}}-[{\displaystyle \frac{2}{3}}:{\displaystyle \frac{2}{5}}-(3+{\displaystyle \frac{1}{2}})]\cdot {\displaystyle \frac{3}{11}}=$$ Empezamos por el corchete, hacemos la división y después el paréntesis:

$$={\displaystyle \frac{5}{3}}-[{\displaystyle \frac{2\cdot 5}{3\cdot 2}}-({\displaystyle \frac{6}{2}}+{\displaystyle \frac{1}{2}})]\cdot {\displaystyle \frac{3}{11}}={\displaystyle \frac{5}{3}}-[{\displaystyle \frac{10}{6}}-\left({\displaystyle \frac{7}{2}}\right)]\cdot {\displaystyle \frac{3}{11}}={\displaystyle \frac{5}{3}}-[{\displaystyle \frac{10}{6}}-\left({\displaystyle \frac{21}{6}}\right)]\cdot {\displaystyle \frac{3}{11}}=$$ $$={\displaystyle \frac{5}{3}}-\left[{\displaystyle \frac{-11}{6}}\right]\cdot {\displaystyle \frac{3}{11}}={\displaystyle \frac{5}{3}}-\left[{\displaystyle \frac{-\cancel{11}}{6}}\right]\cdot {\displaystyle \frac{3}{\cancel{11}}}={\displaystyle \frac{5}{3}}-\left[{\displaystyle \frac{-3}{6}}\right]={\displaystyle \frac{10}{6}}+{\displaystyle \frac{3}{6}}={\displaystyle \frac{13}{6}}$$ Ahora vamos con el denominador:

$${\displaystyle \frac{14}{3}}-{\displaystyle \frac{13}{3}}:({\displaystyle \frac{2}{5}}-3)+{\displaystyle \frac{1}{2}}=$$ Hacemos el paréntesis, luego la división y después el resto de operaciones de izquierda a derecha:

$${\displaystyle \frac{14}{3}}-{\displaystyle \frac{13}{3}}:({\displaystyle \frac{2}{5}}-{\displaystyle \frac{15}{5}})+{\displaystyle \frac{1}{2}}={\displaystyle \frac{14}{3}}-{\displaystyle \frac{13}{3}}:(-{\displaystyle \frac{13}{5}})+{\displaystyle \frac{1}{2}}={\displaystyle \frac{14}{3}}+{\displaystyle \frac{13\cdot 5}{3\cdot 13}}+{\displaystyle \frac{1}{2}}={\displaystyle \frac{14}{3}}+{\displaystyle \frac{\cancel{13}\cdot 5}{3\cdot \cancel{13}}}+{\displaystyle \frac{1}{2}}=$$ $$={\displaystyle \frac{14}{3}}+{\displaystyle \frac{5}{3}}+{\displaystyle \frac{1}{2}}={\displaystyle \frac{19}{3}}+{\displaystyle \frac{1}{2}}={\displaystyle \frac{38}{6}}+{\displaystyle \frac{3}{6}}={\displaystyle \frac{41}{6}}$$ Sustituimos en (*) y tenemos:

$${\displaystyle \frac{{\displaystyle \frac{5}{3}}-[{\displaystyle \frac{2}{3}}:{\displaystyle \frac{2}{5}}-(3+{\displaystyle \frac{1}{2}})]\cdot {\displaystyle \frac{3}{11}}}{{\displaystyle \frac{14}{3}}-{\displaystyle \frac{13}{3}}:({\displaystyle \frac{2}{5}}-3)+{\displaystyle \frac{1}{2}}}}={\displaystyle \frac{{\displaystyle \frac{13}{6}}}{{\displaystyle \frac{41}{6}}}}={\displaystyle \frac{13\cdot 6}{41\cdot 6}}={\displaystyle \frac{13\cdot \cancel{6}}{41\cdot \cancel{6}}}={\displaystyle \frac{13}{41}}$$

Iremos actualizando esta sección con los ejercicios que nos pidáis, para ello podéis mandar un correo a profesor.maties@gmail.com