Đáp án:

`a)`

$\left\{\begin{matrix}\dfrac{2x+2}{3y-1}-\dfrac{3y-2}{x-1}=2\\ \dfrac{-x-1}{3y-1}-\dfrac{3y-2}{2x-2}\end{matrix}\right.$

$⇔\left\{\begin{matrix}\dfrac{2(x+1)}{3y-1}-\dfrac{3y-2}{x-1}=2\\ \dfrac{-(x+1)}{3y-1}-\dfrac{3y-2}{2(x-1)}\end{matrix}\right.$

Đặt `(x+1)/(3y-1)=a`    (`y \ne 1/3`)

`(3y-2)/(x-1)=b`   (`x \ne 1`).

Ta có :

$\left\{\begin{matrix}2a-b=2\\ -a-\dfrac{1}{2}b=-2\end{matrix}\right.$

$⇔\left\{\begin{matrix}a-\dfrac{1}{2}b=1\\ -a-\dfrac{1}{2}b=-2\end{matrix}\right.$

$⇔\left\{\begin{matrix}a-\dfrac{1}{2}b=1\\ -a-\dfrac{1}{2}b=-2\end{matrix}\right.$

$⇔\left\{\begin{matrix}2a=3\\ 2a-b=2\end{matrix}\right.$

$⇔\left\{\begin{matrix}a=\dfrac{3}{2}\\ 2.\dfrac{3}{2}-b=2\end{matrix}\right.$

$⇔\left\{\begin{matrix}a=\dfrac{3}{2}\\ b=1\end{matrix}\right.$

$⇔\left\{\begin{matrix}\dfrac{x+1}{3y-1}=\dfrac{3}{2}\\ \dfrac{3y-2}{x-1}=1\end{matrix}\right.$

$⇔\left\{\begin{matrix}2(x+1)=3(3y-1)\\ 3y-2=x-1\end{matrix}\right.$

$⇔\left\{\begin{matrix}2x+2=9y-3\\ 3y-2=x-1\end{matrix}\right.$

$⇔\left\{\begin{matrix}2x-9y=-5\\ -x+3y=1\end{matrix}\right.$

$⇔\left\{\begin{matrix}2x-9y=-5\\ -3x+9y=3\end{matrix}\right.$

$⇔\left\{\begin{matrix}-x=-2\\ -x+3y=1\end{matrix}\right.$

$⇔\left\{\begin{matrix}x=2\\ -2+3y=1\end{matrix}\right.$

$⇔\left\{\begin{matrix}x=2\\ y=1 \end{matrix}\right.$

Vậy hệ phương trình có nghiệm duy nhất `(x;y)=(2;1)`

`b)`