giải pt

V

vipboycodon

$\dfrac{2}{x^3-x^2-x+1} = \dfrac{3}{1-x^2}-\dfrac{1}{x+1}$
Đk : $x \ne \pm 1$
<=> $\dfrac{2}{x^2(x-1)-(x-1)} = \dfrac{-3}{(x-1)(x+1)}-\dfrac{1}{x+1}$
<=> $\dfrac{2}{(x-1)^2(x+1)} = \dfrac{-3}{(x-1)(x+1)}-\dfrac{1}{x+1}$
<=> $2 = -3(x-1)-(x-1)^2$
<=> $2 = -3x+3-(x^2-2x+1)$
<=> $x^2+x = 0$
<=> $x = 0$ hoặc $x = -1$
 
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