a, Let's \(\dfrac{1}{x+1}=a;\dfrac{1}{y-1}=b\), we have:
\(\left\{{}\begin{matrix}3a+b=2\\2a-3b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2-3a\\2a-3\left(2-3a\right)=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2-3a\\2a-6+9a=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2-3a\\11a=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2-3\cdot1\\a=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=-1\\a=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=-1\\\dfrac{1}{y-1}=1\end{matrix}\right.\)(remember \(\left(x;y\right)\ne-1;1\) :>)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=-1\\y-1=1\end{matrix}\right.\\ \left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\) (satisfied)
So equations (i don't know word "hệ phương trình" in English :>) have 1 root \(\left(x;y\right)=\left(-2;2\right)\).
Enjoy
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a: \(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{1}{y-1}=2\\\dfrac{2}{x+1}-\dfrac{3}{y-1}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{x+1}+\dfrac{3}{y-1}=6\\\dfrac{2}{x+1}-\dfrac{3}{y-1}=5\end{matrix}\right.\)
=>11/x+1=11 và 1/y-1=2-3/x+1
=>x+1=1 và 1/y-1=2-3=-1
=>x=0; y-1=-1
=>x=0; y=0
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+1}-\dfrac{6}{y-1}=2\\\dfrac{2}{x+1}+\dfrac{4}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{10}{y-1}=-1\\\dfrac{1}{x+1}=1+\dfrac{3}{y-1}\end{matrix}\right.\)
=>y-1=10; 1/x+1=1+3/10=13/10
=>y=11; x=10/13-1=-3/13
