Đáp án: $x=\dfrac{11\pm\sqrt{65}}{4}$
Giải thích các bước giải:
ĐKXĐ: $x\ne \dfrac32, x\ne -1$
Nếu $x>-1\to x+1>0\to |x+1|=x+1$
$\dfrac{x-1}{2x-3}=\dfrac{-3x+1}{|x+1|}$
$\to \dfrac{x-1}{2x-3}=\dfrac{-3x+1}{x+1}$
$\to \left(x-1\right)\left(x+1\right)=\left(2x-3\right)\left(-3x+1\right)$
$\to -7x^2+11x-2=0$
$\to x=\dfrac{11\pm\sqrt{65}}{4}$ (chọn)
Nếu $x<-1\to x+1<0\to |x+1|=-(x+1)$
$\to \dfrac{x-1}{2x-3}=\dfrac{-3x+1}{|x+1|}$
$\to \dfrac{x-1}{2x-3}=\dfrac{-3x+1}{-(x+1)}$
$\to \left(x-1\right)\left(x+1\right)=-\left(2x-3\right)\left(-3x+1\right)$
$\to 5x^2-11x+4=0$
$\to x=\dfrac{11+\pm\sqrt{41}}{10}$ (loại vì $x<-1$)
