Lấy PT1 trừ PT2 ta được
\(\sqrt{4x+y}-\frac{5}{3}x+\frac{1}{6}y=3\)
\(\Leftrightarrow6\sqrt{4x+y}-10x+y=18\)
đặt \(\sqrt{4x+y}=a\left(a\ge0\right)\)
\(\Rightarrow6a-\frac{5a^2-7y}{2}=18\)
\(\Leftrightarrow12a-5a^2+7y=36\)
Giải a theo y, rồi thay vào
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{4x+y}=a\ge0\\\sqrt{x+2y}=b\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{2a^2-b^2}{7}\\y=\frac{4b^2-a^2}{7}\end{matrix}\right.\)
Ta được: \(\left\{{}\begin{matrix}a+b=5\\\frac{5\left(2a^2-b^2\right)}{21}-\frac{4b^2-a^2}{42}+b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\21a^2-14b^2+42b-84=0\end{matrix}\right.\)
\(\Rightarrow21\left(5-b\right)^2-14b^2+42b-84=0\)
\(\Leftrightarrow b^2-24b+63=0\Rightarrow\left[{}\begin{matrix}b=21\Rightarrow a=-16\left(l\right)\\b=3\Rightarrow a=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{4x+y}=2\\\sqrt{x+2y}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x+y=4\\x+2y=9\end{matrix}\right.\) \(\Leftrightarrow...\)
