Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=a\\\dfrac{1}{x-y}=b\end{matrix}\right.\)
Hệ: \(\left\{{}\begin{matrix}a-2b=2\\5-4b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{3}\\b=-\dfrac{7}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-3\\x-y=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{27}{14}\\x=-\dfrac{15}{14}\end{matrix}\right.\)
Vậy....
Ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x+y}-\dfrac{2}{x-y}=2\\\dfrac{5}{x+y}-\dfrac{4}{x-y}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+y}-\dfrac{10}{x-y}=10\\\dfrac{5}{x+y}-\dfrac{4}{x-y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-14}{x-y}=7\\\dfrac{1}{x+y}-\dfrac{2}{x-y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-2\\\dfrac{1}{x+y}-\dfrac{2}{-2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-2\\\dfrac{1}{x+y}+1=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-2\\\dfrac{1}{x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-2\\x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=1-x=1-\dfrac{-1}{2}=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\)
