Giải thích các bước giải:
ĐKXĐ : $x\ne\pm1$
Ta có :
$\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}=3x(1-\dfrac{x-1}{x+1})$
$\to \dfrac{x+1}{x-1}\left(x-1\right)\left(x+1\right)+\dfrac{x-1}{x+1}\left(x-1\right)\left(x+1\right)=3x\left(1-\dfrac{x-1}{x+1}\right)\left(x-1\right)\left(x+1\right)$
$\to \left(x+1\right)^2+\left(x-1\right)^2=3x((x-1)(x+1)-(x-1)^2)$
$\to 2x^2+2=6x^2-6x$
$\to 4x^2-6x-2=0$
$\to 2x^2-3x-1=0$
$\to 2\left(x-\dfrac{3}{4}\right)^2-\dfrac{17}{8}=0$
$\to \left(x-\dfrac{3}{4}\right)^2=\dfrac{17}{16}$
$\to x=-\dfrac{-3+\sqrt{17}}{4},\:x=\dfrac{3+\sqrt{17}}{4}$
