ĐKXĐ: \(\left\{{}\begin{matrix}x+y\ne0\\x-y\ne0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\dfrac{1}{x-y}=v\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}2u+v=3\\u-3v=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6u+3v=9\\u-3v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7u=10\\u-3v=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=\dfrac{10}{7}\\v=\dfrac{1}{7}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y}=\dfrac{10}{7}\\\dfrac{1}{x-y}=\dfrac{1}{7}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=\dfrac{7}{10}\\x-y=7\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{77}{20}\\y=-\dfrac{63}{20}\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{2}{x+y}+\dfrac{1}{x-y}=3\\\dfrac{1}{x+y}-\dfrac{3}{x-y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+y}+\dfrac{1}{x-y}=3\\\dfrac{2}{x+y}-\dfrac{6}{x-y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{x-y}=1\\\dfrac{1}{x+y}-\dfrac{3}{x-y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=7\\\dfrac{1}{x+y}=1+\dfrac{3}{x-y}=\dfrac{10}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=7\\x+y=\dfrac{7}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{77}{10}\\x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{77}{20}\\y=\dfrac{-63}{20}\end{matrix}\right.\)
