Đáp án: a.$x\in\{\dfrac52,-1\}$
b.$x=-2$
Giải thích các bước giải:
a.ĐKXĐ: $x\ne 1,-2$
Ta có :
$\dfrac{2}{x-1}=\dfrac{1+2x}{x+2}$
$\to 2\left(x+2\right)=\left(x-1\right)\left(1+2x\right)$
$\to 2x+4=2x^2-x-1$
$\to 2x^2-3x-5=0$
$\to 2x^2+2x-5x-5=0$
$\to 2x(x+1)-5(x+1)=0$
$\to (2x-5)(x+1)=0$
$\to x\in\{\dfrac52,-1\}$
b.ĐKXĐ $x\ne 0,2,-1$
Ta có :
$\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-x-2}$
$\to \dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{(x-2)(x+1)}$
$\to x\left(x+2\right)\left(x+1\right)-\left(x-2\right)\left(x+1\right)=2x$
$\to x^3+2x^2+3x+2=2x$
$\to x^3+2x^2+x+2=0$
$\to (x^3+x)+(2x^2+2)=0$
$\to x(x^2+1)+2(x^2+1)=0$
$\to (x+2)(x^2+1)=0$
$\to x+2=0$ vì $x^2+1>0$
$\to x=-2$
