ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+1\right)-y\left(x+1\right)=15\\15\left(\frac{1}{x+1}+\frac{1}{y-2}\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y-2\right)=-15\\15\left(\frac{1}{x+1}+\frac{1}{y-2}\right)=2\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+1=a\\y-2=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}ab=-15\\15\left(\frac{1}{a}+\frac{1}{b}\right)=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ab=-15\\\frac{15\left(a+b\right)}{ab}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}ab=-15\\a+b=-2\end{matrix}\right.\)
Theo Viet đảo, a và b là nghiệm:
\(t^2+2t-15=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1=3\\y-2=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-5\\y-2=3\end{matrix}\right.\end{matrix}\right.\)