Question

como resolver: $\frac{x}{x-1}$ + $\frac{2}{ x^{2} -1}$ - $\frac{8}{x+1}$ = 0 ?

Answer 1

$\dfrac{x}{x-1}+\dfrac{2}{x^2-1}-\dfrac{8}{x+1}=0\\\\\\ \dfrac{x}{x-1}+\dfrac{2}{(x-1)(x+1)}-\dfrac{8}{x+1}=0$


Reduzindo os termos ao mesmo denominador,

$\dfrac{x(x+1)}{(x-1)(x+1)}+\dfrac{2}{(x-1)(x+1)}-\dfrac{8(x-1)}{(x-1)(x+1)}=0\\\\\\ \dfrac{x(x+1)+2-8(x-1)}{(x-1)(x+1)}=0\\\\\\ \dfrac{x^2+x+2-8x+8}{(x-1)(x+1)}=0\\\\\\ \dfrac{x^2-7x+10}{(x-1)(x+1)}=0$


Como o denominador deve ser diferente de zero, isto é,

$x\ne 1$  e  $x\ne -1,$


para que a fração seja igual a zero, basta que o numerador se anule:

$x^2-7x+10=0~~~\Rightarrow~~\left\{ \!\begin{array}{l} a=1\\b=-7\\c=10 \end{array} \right.$


$\Delta=b^2-4ac\\\\ \Delta=(-7)^2-4\cdot 1\cdot 10\\\\ \Delta=49-40\\\\ \Delta=9\\\\\\\\ x=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ x=\dfrac{-(-7)\pm \sqrt{9}}{2\cdot 1}\\\\\\ x=\dfrac{7\pm 3}{2}$


$\\\begin{array}{rcl} x=\dfrac{7-3}{2}&~\text{ ou }~&x=\dfrac{7+3}{2}\\\\\\ x=\dfrac{4}{2}&~\text{ ou }~&x=\dfrac{10}{2} \end{array}\\\\\\\\ ~~~~~~\boxed{ \begin{array}{rcl} x=2&~\text{ ou }~&x=5 \end{array}}$



Conjunto solução:   S = {2, 5}.


Dúvidas? Comente.


Bons estudos! :-)