Expresiones algebraicas: Operaciones y simplificación. Ejercicios resueltos.

Operaciones con expresiones algebraicas. Simplificar. Ejercicios resueltos.

Usar que $x^3-y^3=(x-y)\cdot (x^2+xy+y^2)$ y $x^3+y^3=(x+y)\cdot (x^2-xy+y^2)$

Sustituimos ${\displaystyle \frac{x^3-y^3}{x-y}}$ por $x^2+xy+y^2$ y $x^3+y^3=(x+y)\cdot (x^2-xy+y^2)$:
$${\displaystyle \frac{x^3-y^3}{x-y}}\cdot {\displaystyle \frac{x+y}{x^3+y^3+2xy(x+y)}}={\displaystyle \frac{(x^2+xy+y^2)\cdot (x+y)}{(x+y)\cdot (x^2-xy+y^2)+2xy(x+y)}}=$$
Sacamos factor común en el denominador a $x+y$ tenemos que:

$$={\displaystyle \frac{(x^2+xy+y^2)\cdot (x+y)}{(x+y)\cdot (x^2-xy+y^2+2xy)}}={\displaystyle \frac{(x^2+xy+y^2)\cdot (x+y)}{(x+y)\cdot (x^2+xy+y^2)}}=1$$

Vayamos por partes:

$1-{\displaystyle \frac{m}{m+1}}={\displaystyle \frac{m+1-m}{m+1}}={\displaystyle \frac{1}{m+1}}$

$1+{\displaystyle \frac{m}{m+1}}={\displaystyle \frac{m+1+m}{m+1}}={\displaystyle \frac{2m+1}{m+1}}$

${\displaystyle \frac{2}{m}}+{\displaystyle \frac{1}{m^2}}={\displaystyle \frac{2m+1}{m^2}}$

$1-{\displaystyle \frac{1}{m}}={\displaystyle \frac{m-1}{m}}$

Así:

$${\displaystyle \frac{{(1-{\displaystyle \frac{m}{m+1}})}^2}{1+{\displaystyle \frac{m}{m+1}}}}\cdot {\displaystyle \frac{{\displaystyle \frac{2}{m}}+{\displaystyle \frac{1}{m^2}}}{1-{\displaystyle \frac{1}{m}}}}={\displaystyle \frac{{\displaystyle \frac{1}{(m+1{)}^2}}\cdot {\displaystyle \frac{2m+1}{m^2}}}{{\displaystyle \frac{2m+1}{m+1}}\cdot {\displaystyle \frac{m-1}{m}}}}={\displaystyle \frac{{\displaystyle \frac{1}{(m+1{)}^{\cancel{2}}}}\cdot {\displaystyle \frac{\cancel{2m+1}}{m^{\cancel{2}}}}}{{\displaystyle \frac{\cancel{2m+1}}{\cancel{m+1}}}\cdot {\displaystyle \frac{m-1}{\cancel{m}}}}}={\displaystyle \frac{{\displaystyle \frac{1}{m+1}}\cdot {\displaystyle \frac{1}{m}}}{m-1}}={\displaystyle \frac{1}{m\cdot (m^2-1)}}$$

Vamos por partes, 1º:

${\displaystyle \frac{1+x}{1-x}}-{\displaystyle \frac{1-x}{1+x}}={\displaystyle \frac{(1+x{)}^2}{(1-x)\cdot (1+x)}}-{\displaystyle \frac{(1-x{)}^2}{(1+x)\cdot (1-x)}}={\displaystyle \frac{(1+x{)}^2-(1-x{)}^2}{(1+x)\cdot (1-x)}}={\displaystyle \frac{1+2x+x^2-1+2x-x^2}{(1+x)\cdot (1-x)}}=$

$={\displaystyle \frac{4x}{(1+x)\cdot (1-x)}}$

Ahora hacemos:

${\displaystyle \frac{1+x}{1-x}}-1={\displaystyle \frac{1+x}{1-x}}-{\displaystyle \frac{1-x}{1-x}}={\displaystyle \frac{1+x-1+x}{1-x}}={\displaystyle \frac{2x}{1-x}}$

Y por último:

$1-{\displaystyle \frac{1}{1+x}}={\displaystyle \frac{1+x}{1+x}}-{\displaystyle \frac{1}{1+x}}={\displaystyle \frac{1+x-1}{1+x}}={\displaystyle \frac{x}{1+x}}$

Juntando todo tenemos:

$({\displaystyle \frac{1+x}{1-x}}-{\displaystyle \frac{1-x}{1+x}}):[({\displaystyle \frac{1+x}{1-x}}-1)\cdot (1-{\displaystyle \frac{1}{1+x}})]={\displaystyle \frac{4x}{(1+x)\cdot (1-x)}}:[{\displaystyle \frac{2x}{1-x}}\cdot {\displaystyle \frac{x}{1+x}}]=$

$={\displaystyle \frac{4x}{(1+x)\cdot (1-x)}}:{\displaystyle \frac{2x^2}{(1+x)\cdot (1-x)}}={\displaystyle \frac{4x}{2x^2}}={\displaystyle \frac{2}{x}}$

Vamos con la primera fracción:
$${\displaystyle \frac{{\displaystyle \frac{2x}{x-y}}}{{\displaystyle \frac{4x}{x^2+2xy+y^2}}}}={\displaystyle \frac{2x(x+y{)}^2}{4x(x-y)}}={\displaystyle \frac{(x+y{)}^2}{2(x-y)}}$$
Vamos con la segunda fracción:
$${\displaystyle \frac{x(x^2-y^2)}{x+y}}={\displaystyle \frac{x(x+y)(x-y)}{x+y}}=x(x-y)$$
Ahora elevamos al cuadrado:
$${[{\displaystyle \frac{{\displaystyle \frac{2x}{x-y}}}{{\displaystyle \frac{4x}{x^2+2xy+y^2}}}}:{\displaystyle \frac{{\displaystyle \frac{x}{x+y}}}{{\displaystyle \frac{1}{x^2-y^2}}}}]}^2={[{\displaystyle \frac{(x+y{)}^2}{2(x-y)}}:x(x-y)]}^2={\left[{\displaystyle \frac{(x+y{)}^2}{2x(x-y{)}^2}}\right]}^2={\displaystyle \frac{(x+y{)}^4}{4x^2(x-y{)}^4}}$$
Vamos con la tercera fracción, primero el numerador:

$1+{\displaystyle \frac{y}{x}}={\displaystyle \frac{x+y}{x}}$ y ahora el denominador: $1-{\displaystyle \frac{y}{x}}={\displaystyle \frac{x-y}{x}}$. Juntando ambos:

$${\displaystyle \frac{1+{\displaystyle \frac{y}{x}}}{1-{\displaystyle \frac{y}{x}}}}={\displaystyle \frac{{\displaystyle \frac{x+y}{x}}}{{\displaystyle \frac{x-y}{x}}}}={\displaystyle \frac{x+y}{x-y}}$$
Elevamos a la cuarta y tenemos:
$${\left[{\displaystyle \frac{x+y}{x-y}}\right]}^4={\displaystyle \frac{(x+y{)}^4}{(x-y{)}^4}}$$
Juntando todo tenemos:
$${\displaystyle \frac{(x+y{)}^4}{4x^2(x-y{)}^4}}:{\displaystyle \frac{(x+y{)}^4}{(x-y{)}^4}}={\displaystyle \frac{(x+y{)}^4(x-y{)}^4}{4x^2(x-y{)}^4(x+y{)}^4}}={\displaystyle \frac{\cancel{(x+y{)}^4}\xcancel{(x-y{)}^4}}{4x^2\xcancel{(x-y{)}^4}\cancel{(x+y{)}^4}}}={\displaystyle \frac{1}{4x^2}}$$

Vamos con la primera fracción, con el denominador:

${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}={\displaystyle \frac{x+y}{xy}}$ cogemos el numerador y tenemos:

$${\displaystyle \frac{x^2-y^2}{{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}}}={\displaystyle \frac{x^2-y^2}{{\displaystyle \frac{x+y}{xy}}}}={\displaystyle \frac{xy(x^2-y^2)}{x+y}}={\displaystyle \frac{xy(x-y)(x+y)}{x+y}}=xy(x-y)$$
Vamos con la segunda fracción:

con el numerador $x-{\displaystyle \frac{x^2}{x+y}}={\displaystyle \frac{x^2+xy-x^2}{x+y}}={\displaystyle \frac{xy}{x+y}}$

cogemos el denominador y tenemos: $y-{\displaystyle \frac{y^2}{x+y}}={\displaystyle \frac{xy+y^2-y^2}{x+y}}={\displaystyle \frac{xy}{x+y}}$

juntando ambas tenemos:

$${\displaystyle \frac{x-{\displaystyle \frac{x^2}{x+y}}}{y-{\displaystyle \frac{y^2}{x+y}}}}={\displaystyle \frac{{\displaystyle \frac{xy}{x+y}}}{{\displaystyle \frac{xy}{x+y}}}}=1$$ Vamos con la tercera fracción:

con el numerador: ${\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{y}}={\displaystyle \frac{y-x}{xy}}$ y el denominador lo hemos hecho antes: ${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}={\displaystyle \frac{x+y}{xy}}$

juntando tenemos que
$${\displaystyle \frac{{\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{y}}}{{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}}}={\displaystyle \frac{{\displaystyle \frac{y-x}{xy}}}{{\displaystyle \frac{x+y}{xy}}}}={\displaystyle \frac{y-x}{x+y}}$$
Juntando todo tenemos:

$${\displaystyle \frac{x^2-y^2}{{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}}}\cdot {\displaystyle \frac{x-{\displaystyle \frac{x^2}{x+y}}}{y-{\displaystyle \frac{y^2}{x+y}}}}\cdot {\displaystyle \frac{{\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{y}}}{{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}}}=xy(x-y)\cdot 1\cdot {\displaystyle \frac{y-x}{x+y}}={\displaystyle \frac{-xy(x-y{)}^2}{x+y}}$$

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