ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne1\end{matrix}\right.\)
Đặt \(\frac{x}{x+1}=a\) và \(\frac{y}{y-1}=b\)
PT\(\Leftrightarrow\left\{{}\begin{matrix}2a+3b=1\\3a-4b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1-3b}{2}\\3\cdot\frac{1-3b}{2}-4b=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1-3b}{2}\\\frac{3-9b}{2}-\frac{8b}{2}=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1-3b}{2}\\3-9b-8b=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1-3b}{2}\\3-17b=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1-3b}{2}\\-17b=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1-3\cdot\left(-1\right)}{2}=2\\b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{x+1}=2\\\frac{y}{y-1}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\left(x+1\right)\\y=-1\cdot\left(y-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2x+1\\y=1-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2x=1\\y+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\left(loại\right)\\y=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\varnothing\)
đặt a=\(\frac{x}{x+1}\) , b=\(\frac{y}{y+1}\)
\(\Rightarrow\)\(\left\{{}\begin{matrix}2a+3b=1\\3a-4b=10\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6a+9b=3\\6a-8b=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}17b=-17\\6a-8b=20\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=-1\\6a-8\cdot\left(-1\right)=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-1\\6a=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=-1\\a=2\end{matrix}\right.\)
ta có: a=2=\(\frac{x}{x+1}\); b=-1=\(\frac{y}{y+1}\)
\(\Rightarrow\)x=-2 ; y=\(\frac{-1}{2}\)
