Question

\(\left\{{}\begin{matrix}4\sqrt{x-2y}+\dfrac{1}{y-3}=5\\\sqrt{x-2y}+\dfrac{2}{3-y}=-1\end{matrix}\right.\)

Answer 1

ĐKXĐ: \(x-2y\ge0\)

\(\left\{{}\begin{matrix}4\sqrt{x-2y}+\dfrac{1}{y-3}=5\\\sqrt{x-2y}-\dfrac{2}{y-3}=-1\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-2y}=u\ge0\\\dfrac{1}{y-3}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4u+v=5\\u-2v=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2y}=1\\\dfrac{1}{y-3}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-2y=1\\y-3=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=4\\x=2y+1=9\end{matrix}\right.\)