Question

AYUDAAA NO ENTIENDO NADA

Answer 1

Estas son las operaciones y las simplificaciones de los ejercicios planteados

6.

a) $\frac{\frac{4}{x}-x}{\frac{1}{x}+\frac{1}{2}}=-2\left(x-2\right)$

b) $\frac{x+2}{\left(x+2\right)^2}\cdot \frac{x^2-4}{x}=\frac{x-2}{x}$

7.

a)  $\left(\frac{x+y}{x-y}-\frac{x-y}{x+y}\right)\left(\frac{x}{y}-\frac{y}{x}\right)=4$

8.

a) $\frac{1}{x-1}+\frac{1}{x-3}-\frac{x-1}{x^2-4x+3}=\frac{1}{x-1}$

9.

a)$\frac{\frac{9+6x+x^2}{9-x^2}\cdot \frac{3x^2-x^3}{3x^2+x^3}}{\frac{\frac{2x-4x}{\frac{3}{4}-\frac{2}{8}}}{\frac{2x^2-8x+8}{x-2}}}=-\frac{x-2}{2x}$

Resolución paso a paso

6. Opera y Simplifica

a)  

$\left(\frac{4}{x}-x\right)\div \left(\frac{1}{x}+\frac{1}{2}\right)$

$\left(\frac{4}{x}-x\right)\div \left(\frac{1}{x}+\frac{1}{2}\right)=\frac{\frac{4}{x}-x}{\frac{1}{x}+\frac{1}{2}}$

Simplificando Fracción

$=\frac{\frac{4}{x}-x}{\frac{2+x}{2x}}=\frac{\frac{4-x^2}{x}}{\frac{2+x}{2x}}$

Dividir Fracciones : $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot \:d}{b\cdot \:c}$

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

$\mathrm{Factorizar}\:2\left(-x^2+4\right)$  

$2\left(-x^2+4\right)=-2\left(x+2\right)\left(x-2\right)$

$=-\frac{2\left(x+2\right)\left(x-2\right)}{2+x}$

Simplificando

$=-2\left(x-2\right)$

entonces

$\frac{\frac{4}{x}-x}{\frac{1}{x}+\frac{1}{2}}=-2\left(x-2\right)$

b)

$\frac{x+2}{\left(x+2\right)^2}\cdot \frac{x^2-4}{x}$

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

Multiplicacion de Fracciones:$\frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d}$

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

$\mathrm{Factorizar}\:x^2-4$

$x^2-4=\left(x+2\right)\left(x-2\right)$

$=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)x}$

Eliminar términos Comunes

$=\frac{x-2}{x}$

Entonces

$\frac{x+2}{\left(x+2\right)^2}\cdot \frac{x^2-4}{x}=\frac{x-2}{x}$

7.

a)

$\left(\frac{x+y}{\:x-y}-\frac{x-y}{x+y}\right)\cdot \left(\frac{x}{y}-\frac{y}{x}\right)$

$\mathrm{Simplificar}\:\frac{x+y}{x-y}-\frac{x-y}{x+y}$

El mínimo común múltiplo de $x-y,\:x+y:\quad \left(x-y\right)\left(x+y\right)$

$=\frac{\left(x+y\right)^2-\left(x-y\right)^2}{\left(x-y\right)\left(x+y\right)}$

$\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-\left(x-y\right)^2=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)=x^2+2xy+y^2-x^2+2xy-y^2\\\\=4xy$

$=\frac{4xy}{\left(x-y\right)\left(x+y\right)}$

$=\frac{4xy}{\left(x-y\right)\left(x+y\right)}\left(\frac{x}{y}-\frac{y}{x}\right)$

$\mathrm{Multiplicar\:fracciones}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}$

$=\frac{4xy\left(\frac{x}{y}-\frac{y}{x}\right)}{\left(x-y\right)\left(x+y\right)}$

$=\frac{4\cdot \frac{x^2-y^2}{xy}xy}{\left(x-y\right)\left(x+y\right)}=\frac{4\left(x+y\right)\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}$

$\frac{4\left(x+y\right)\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{4\left(x-y\right)}{x-y}=4$

entonces

$\left(\frac{x+y}{x-y}-\frac{x-y}{x+y}\right)\left(\frac{x}{y}-\frac{y}{x}\right)=4$

8.

a)

$\frac{1}{x-1}+\frac{1}{x-3}-\frac{x-1}{x^2-4x+3}$

Simplificando el tercer termino

$\frac{x-1}{x^2-4x+3}=\frac{x-1}{\left(x-1\right)\left(x-3\right)}$

Eliminando Términos Comunes  

$=\frac{x-3}{\left(x-1\right)\left(x-3\right)}+\frac{0}{\left(x-1\right)\left(x-3\right)}$

$\mathrm{Ya\:que\:los\:denominadores\:son\:iguales,\:combinar\:las\:fracciones}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{x-3+0}{\left(x-1\right)\left(x-3\right)}=\frac{x-3}{\left(x-1\right)\left(x-3\right)}$

$=\frac{x-3}{\left(x-1\right)\left(x-3\right)}$

$\mathrm{Eliminar\:los\:terminos\:comunes:}\:x-3$

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

Entonces

$\frac{1}{x-1}+\frac{1}{x-3}-\frac{x-1}{x^2-4x+3}=\frac{1}{x-1}$

9.

a)

$\frac{\frac{9+6x+x^2}{9-x^2}\cdot \frac{3x^2-x^3}{3x^2+x^3}}{\frac{\frac{2x-4x}{\frac{3}{4}-\frac{2}{8}}}{\frac{2x^2-8x+8}{x-2}}}$

$=\frac{\frac{x^2+6x+9}{-x^2+9}\cdot \frac{-x^3+3x^2}{x^3+3x^2}}{-\frac{2x\left(x-2\right)}{\left(\frac{3}{4}-\frac{2}{8}\right)\left(2x^2-8x+8\right)}}$

Simplificando  

$=-\frac{\frac{9+6x+x^2}{9-x^2}\cdot \frac{3x^2-x^3}{3x^2+x^3}}{\frac{2x\left(x-2\right)}{\left(\frac{3}{4}-\frac{2}{8}\right)\left(2x^2-8x+8\right)}}$

$=-\frac{\frac{9+6x+x^2}{9-x^2}\cdot \frac{3x^2-x^3}{3x^2+x^3}\left(\frac{3}{4}-\frac{2}{8}\right)\left(2x^2-8x+8\right)}{2x\left(x-2\right)}$

$=-\frac{\left(x-2\right)^2}{2x\left(x-2\right)}$

$=-\frac{x-2}{2x}$

entonces

$\frac{\frac{9+6x+x^2}{9-x^2}\cdot \frac{3x^2-x^3}{3x^2+x^3}}{\frac{\frac{2x-4x}{\frac{3}{4}-\frac{2}{8}}}{\frac{2x^2-8x+8}{x-2}}}=-\frac{x-2}{2x}$