\(\left\{{}\begin{matrix}2\left(x-1\right)+\dfrac{3}{y+2}=5\\\left(x-1\right)-\dfrac{1}{y+2}=\dfrac{5}{3}\end{matrix}\right.\) \(\left(dk:y\ne-2\right)\)
Đặt \(\dfrac{1}{y+2}=a\), Hệ pt trở thành : \(\left\{{}\begin{matrix}2\left(x-1\right)+3a=5\\\left(x-1\right)-a=\dfrac{5}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)+3a=5\\2\left(x-1\right)-2a=\dfrac{10}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)+3a=5\\3a+2a=5-\dfrac{10}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)+3a=5\left(1\right)\\5a=\dfrac{5}{3}\end{matrix}\right.\)
\(\Leftrightarrow a=\dfrac{1}{3}\)
Ta có : \(\dfrac{1}{y+2}=\dfrac{1}{3}\Rightarrow y+2=3\Rightarrow y=1\left(tmdk\right)\)
Thay \(a=\dfrac{1}{3}\) vào \(\left(1\right)\Rightarrow2x-2+3.\dfrac{1}{3}=5\Rightarrow2x=6\Rightarrow x=3\)
Vậy hệ pt có nghiệm duy nhất \(\left(x;y\right)=\left(3;1\right)\)