Đáp án: Vô nghiệm
Giải thích các bước giải:
ĐKXĐ: $x\ne \pm1,y\ne \pm1$
Ta có:
$\dfrac{x}{x-1}+\dfrac{2y}{y+1}=3$
$\to (\dfrac{x}{x-1}-1)+(\dfrac{2y}{y+1}-2)=0$
$\to \dfrac{x-(x-1)}{x-1}+\dfrac{2y-2(y+1)}{y+1}=0$
$\to \dfrac{1}{x-1}+\dfrac{-2}{y+1}=0$
$\to \dfrac{1}{x-1}=\dfrac{2}{y+1}$
$\to y+1=2(x-1)$
$\to y+1=2x-2$
$\to y=2x-3$
Mà $\dfrac{x}{x+1}+\dfrac{2y}{y-1}=2$
$\to\dfrac{x}{x+1}+\dfrac{2y}{y-1}-2=0$
$\to\dfrac{x}{x+1}+\dfrac{2y-2(y-1)}{y-1}=0$
$\to\dfrac{x}{x+1}+\dfrac{2}{y-1}=0$
$\to\dfrac{x}{x+1}+\dfrac{2}{2x-3-1}=0$
$\to\dfrac{x}{x+1}+\dfrac{2}{2x-4}=0$
$\to\dfrac{x}{x+1}+\dfrac{1}{x-2}=0$
$\to x\left(x-2\right)+x+1=0$
$\to x^2-x+1=0$
$\to (x-\dfrac12)^2+\dfrac34=0$
$\to$Phương trình vô nghiệm
