\(\left\{{}\begin{matrix}2\sqrt{x-3}+\frac{12}{y-2x}=8\\3\sqrt{4x-12}+\frac{3}{2x-y}=\frac{9}{2}\end{matrix}\right.\) \(Đkxđ:\left\{{}\begin{matrix}x\ge3\\y\ne2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-3}+\frac{12}{y-2x}=8\\6\sqrt{x-3}+\frac{3}{2x-y}=\frac{9}{2}\end{matrix}\right.\)
Đặt: \(\left\{{}\begin{matrix}2\sqrt{x-3}=a\left(a>0\right)\\\frac{3}{2x-y}=b\end{matrix}\right.\)
Ta được phương trình mới:
\(\left\{{}\begin{matrix}a-4b=8\\3a+b=\frac{9}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-3}=2\\\frac{3}{2x-y}=-\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=1\\2x-y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=10\end{matrix}\right.\)
Vậy ..........
