Đáp án:
$\left[ \begin{array}{l}x=\dfrac{\sqrt{2}-2}{2}\\x=\dfrac{-\sqrt{2}-2}{2}\end{array} \right.$
Giải thích các bước giải:
$\dfrac{x^2-1}{x-1}=\dfrac{A}{2A(x+1)}$
$↔\dfrac{(x-1)(x+1)}{x-1}=\dfrac{1}{2(x+1)}$
$↔x+1=\dfrac{1}{2(x+1)}$
$↔(x+1)^2=\dfrac{1}{2}$
$↔\left[ \begin{array}{l}x+1=\dfrac{\sqrt{2}}{2}\\x+1=-\dfrac{\sqrt{2}}{2}\end{array} \right.$
$↔\left[ \begin{array}{l}x=\dfrac{\sqrt{2}-2}{2}\\x=\dfrac{-\sqrt{2}-2}{2}\end{array} \right.$
vậy $x=\dfrac{\sqrt{2}-2}{2}$ hoặc $x=\dfrac{-\sqrt{2}-2}{2}$
