Giải thích các bước giải:
$\dfrac{x^2-x}{x^2-x+1}-\dfrac{x^2-x+2}{x^2-x-2}=1$
$\to \dfrac{x^2-x}{x^2-x+1}-1=\dfrac{x^2-x+2}{x^2-x-2}$
$\to \dfrac{x^2-x-(x^2-x+1)}{x^2-x+1}=\dfrac{x^2-x+2}{x^2-x-2}$
$\to \dfrac{-1}{x^2-x+1}=\dfrac{x^2-x+2}{x^2-x-2}$
Đặt $x^2-x=a$
$\to\dfrac{-1}{a+1}=\dfrac{a+2}{a-2}$
$\to -1\cdot \left(a-2\right)=\left(a+1\right)\left(a+2\right)$
$\to a^2+4a=0$
$\to a\in\{0,-4\}$
$+)a=0\to x^2-x=0\to x\in\{1,0\}$
$+)a=-4\to x^2-x=-4\to x^2-x+4=0\to (x-\dfrac 12)^2+4-\dfrac 14=0\to $ vô nghiệm
