CHÚC BẠN HỌC TỐT!!!
Trả lời:
ĐKXĐ: $x \neq ±1$
$\dfrac{2x^2}{x^2-1}+\dfrac{1}{x-1}+\dfrac{2}{x+1}=1$
$⇔\dfrac{2x^2}{x^2-1}+\dfrac{x+1}{x^2-1}+\dfrac{2x-2}{x^2-1}=\dfrac{x^2-1}{x^2-1}$
$⇔\dfrac{2x^2+x+1+2x-2}{(x-1)(x+1)}=\dfrac{x^2-1}{(x-1)(x+1)}$
$⇔2x^2+3x-1=x^2-1$
$⇔x^2+3x=0$
$⇔x.(x+3)=0$
$⇔\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$
Vậy $S=\{-3;0\}$.
