Giải thích các bước giải:

a.ĐKXĐ: $x\ne 0, y\ne 0$

$\begin{cases}\dfrac1x+\dfrac1y=2\\ \dfrac3x-\dfrac4y=-1\end{cases}$

$\to \begin{cases}\dfrac1x=2-\dfrac1y\\ 3(2-\dfrac1y)-\dfrac4y=-1\end{cases}$

$\to \begin{cases}\dfrac1x=2-\dfrac1y\\ 6-\dfrac7y=-1\end{cases}$

$\to \begin{cases}\dfrac1x=2-\dfrac1y\\ \dfrac7y=7\end{cases}$

$\to \begin{cases}\dfrac1x=1\\ \dfrac1y=1\end{cases}$

$\to \begin{cases}x=1\\y=1\end{cases}$

b.ĐKXĐ: $x\ne 1,y\ne 2$

$\begin{cases}\dfrac6{x-1}-\dfrac5{y-2}=7\\\dfrac3{x-1}+\dfrac2{y-2}=-1\end{cases}$

$\to \begin{cases}2\cdot \dfrac3{x-1}-\dfrac5{y-2}=7\\\dfrac3{x-1}=-1-\dfrac2{y-2}\end{cases}$

$\to \begin{cases}2\cdot (-1-\dfrac2{y-2})-\dfrac5{y-2}=7\\\dfrac3{x-1}=-1-\dfrac2{y-2}\end{cases}$

$\to \begin{cases}-2-\dfrac9{y-2}=7\\\dfrac3{x-1}=-1-\dfrac2{y-2}\end{cases}$

$\to \begin{cases}-\dfrac9{y-2}=9\\\dfrac3{x-1}=-1-\dfrac2{y-2}\end{cases}$

$\to \begin{cases}y-2=-1\\\dfrac3{x-1}=-1-\dfrac2{-1}\end{cases}$

$\to \begin{cases}y=1\\x=4\end{cases}$

c.ĐKXĐ: $x\ne -y, 2x\ne y$

$\begin{cases}\dfrac2{2x-y}-\dfrac6{x+y}=-1\\ \dfrac2{2x-y}-\dfrac1{x+y}=0\end{cases}$

$\to \begin{cases}\dfrac2{2x-y}-6\cdot \dfrac2{2x-y}=-1\\ \dfrac2{2x-y}=\dfrac1{x+y}\end{cases}$

$\to \begin{cases}\dfrac2{2x-y}- \dfrac{12}{2x-y}=-1\\ \dfrac2{2x-y}=\dfrac1{x+y}\end{cases}$

$\to \begin{cases} - \dfrac{10}{2x-y}=-1\\ \dfrac2{2x-y}=\dfrac1{x+y}\end{cases}$

$\to \begin{cases} 2x-y=10\\ \dfrac2{10}=\dfrac1{x+y}\end{cases}$

$\to \begin{cases} 2x-y=10\\x+y=5\end{cases}$

$\to \begin{cases} 2x-y=10\\x+y+2x-y=10+5\end{cases}$

$\to \begin{cases} 2x-y=10\\3x=15\end{cases}$

$\to \begin{cases} 2\cdot 5-y=10\\x=5\end{cases}$

$\to \begin{cases} y=0\\x=5\end{cases}$