ĐKXĐ:....
Đặt \(\left\{{}\begin{matrix}\frac{1}{x-1}=a\\\frac{1}{y+2}=b\end{matrix}\right.\)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}8a+15b=1\\a+b=\frac{1}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{1}{12}-a\\8a+15\cdot\left(\frac{1}{12}-a\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8a+\frac{5}{4}-15a=1\\b=\frac{1}{12}-a\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{28}\\b=\frac{1}{21}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x-1}=\frac{1}{28}\\\frac{1}{y+2}=\frac{1}{21}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=29\\y=19\end{matrix}\right.\) ( thỏa mãn )
Vậy....