ĐKXĐ : `x \ne -1 , y \ne 2`
\(\left\{ \begin{array}{l}\dfrac2{x+1}+\dfrac6{y-2}=5\\\dfrac3{x+1}-\dfrac4{y-2}=1\end{array} \right.\)
Đặt `a = 1/(x+1)` và `b = 1/(y-2)`
`⇔`\(\left\{ \begin{array}{l}2a+6b=5\\3a-4b=1\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}4a+12b=10\\9a-12b=3\end{array} \right.\)
`⇔`\(\left\{\begin{array}{l}13a=13\\3a-4b=1\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}a=1\\3-4b=1\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}a=1\\b=\dfrac12\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}\dfrac1{x+1}=1\\\dfrac1{y-2}=\dfrac12\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}x+1=1\\y-2=2\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}x=0(TM)\\y=4(TM)\end{array} \right.\)
Vậy `(x,y) = (0,4)`
