Question

Giải hệ phương trình:
\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}+\dfrac{5}{2x+y}=2\\\dfrac{1}{2x-y}-\dfrac{1}{2x+y}=\dfrac{2}{5}\end{matrix}\right.\)

Answer 1

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)

Answer 2

Đặt u = 1/(2x-y) ; v = 1/(2x+y) rồi giải