Đặt : \(\left\{{}\begin{matrix}a=\dfrac{1}{2x-y}\\b=\dfrac{1}{2x+y}\end{matrix}\right.\)
Ta có hệ mới :
\(\left\{{}\begin{matrix}3a+5b=2\\a-b=\dfrac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a+5b=2\\5a-5b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8a=4\\a-b=\dfrac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=a-\dfrac{2}{5}=\dfrac{1}{2}-\dfrac{2}{5}=\dfrac{1}{10}\end{matrix}\right.\)
Trả biến :
\(\left\{{}\begin{matrix}\dfrac{1}{2x-y}=\dfrac{1}{2}\\\dfrac{1}{2x+y}=\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=2\\2x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=12\\2x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm : (x;y)= (3;4)
Đặt u = 1/(2x-y) ; v = 1/(2x+y) rồi giải
