ĐKXĐ: \(\left\{{}\begin{matrix}x\ne y\\x\ne-y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{2}{x-y}+\frac{6}{x+y}=1,1\\\frac{4}{x-y}-\frac{9}{x+y}=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{4}{x-y}+\frac{12}{x+y}=2,2\\\frac{4}{x-y}-\frac{9}{x+y}=0,1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\frac{21}{x+y}=2,1\\\frac{2}{x-y}=1,1-\frac{6}{x+y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=10\\\frac{2}{x-y}=1,1-\frac{6}{x+y}=1,1-\frac{6}{10}=\frac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=10\\x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=3\end{matrix}\right.\)(tm)
Vậy hệ phương trình có nghiệm duy nhất là (x;y) = (7;3)
another way to solve
Đặt \(\left\{{}\begin{matrix}\frac{1}{x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}2a+6b=1,1\\4a-9b=0,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1,1-6b}{2}\\\frac{4\cdot\left(1,1-6b\right)}{2}-9b=0,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{1}{10}\\a=\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x-y}=\frac{1}{4}\\\frac{1}{x+y}=\frac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\x+y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=3\end{matrix}\right.\)
Vậy....
