\(\left\{{}\begin{matrix}\frac{3x}{x+1}+\frac{2}{y+4}=4\\\frac{2x}{x+1}-\frac{5}{y+4}=9\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\frac{x}{x+1}\\b=\frac{1}{y+4}\end{matrix}\right.\)
Thay a và b vào hệ phương trình ta có:
\(\left\{{}\begin{matrix}3a+2b=4\\2a-5b=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6a+4b=8\\6a-15b=27\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19b=-19\\3a+2b=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-1\\3a+2.\left(-1\right)=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-1\\a=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=-1\end{matrix}\right.\)
Ta có:
\(a=\frac{x}{x+1}=2\Leftrightarrow x=2\left(x+1\right)\)
<=> x=2x+2
<=> x=-2
\(b=\frac{1}{y+4}=-1\Leftrightarrow y+4=-1\Leftrightarrow y=-5\)
Vậy hệ phương trình có nghiệm \(\left\{{}\begin{matrix}x=-2\\y=-5\end{matrix}\right.\)