Question

Resolva a equação fracionária
$\frac{x+1}{x-1} + \frac{x-1}{x+1} = \frac{13}{6}$

Answer 1

$\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}=\dfrac{13}{6}$

$\dfrac{(x+1)^2+(x-1)^2}{(x-1)(x+1)}=\dfrac{13}{6}$

$\dfrac{x^2+2x+1+x^2-2x+1}{(x-1)(x+1)}=\dfrac{13}{6}$

$\dfrac{2x^2+2}{(x-1)(x+1)}=\dfrac{13}{6}$

$6(2x^2+2)=13(x-1)(x+1)$

$6x^2+12=13x^2-13$

$13x^2-6x^2=12+13$

$7x^2=25$

$x^2=\dfrac{25}{7}$

$x=\pm\dfrac{5}{\sqrt{7}}$.

Answer 2

$\frac{x + 1}{x - 1} + \frac{x - 1}{x + 1} = \frac{13}{6}$

Vamos resolver o primeiro termo

mmc  = (x - 1)*(x + 1)

$\dfrac{(x + 1)*(x + 1) + (x - 1) * (x - 1)}{(x - 1) * (x + 1)} = \dfrac{13}{6} \\ \\ \\ \dfrac{2x^2 + 2}{x^2 - 1} = \dfrac{13}{6}$

Vamos passa tudo para um só lado da igualdade
$\dfrac{2x^2 + 2}{x^2 - 1} - \dfrac{13}{6}$

mmc  = $(x^2 - 1) *6$


$\dfrac{(2x^2 + 2)}{(x^2 - 1)} - \dfrac{13}{6} \\ \\ \\ \dfrac{-((x - 5)* (x + 5))}{6x^2 - 6} => -\dfrac{x^2 - 25}{6x^2 - 6}$