ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3y}=a\ge0\\\sqrt{5-x-y}=b\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=a^2+2b^2-10\\x=15-a^2-3b^2\end{matrix}\right.\)
Ta được hệ:
\(\left\{{}\begin{matrix}2a+b=7\\3b-\sqrt{2\left(15-a^2-3b^2\right)+a^2+2b^2-13}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2a+b=7\\3b-1=\sqrt{17-a^2-4b^2}\end{matrix}\right.\) \(\left(b\ge\frac{1}{3}\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=7-2a\left(a\le\frac{7}{2}\right)\\9b^2-6b+1=17-a^2-4b^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=7-2a\\a^2+13b^2-6b-16=0\end{matrix}\right.\)
\(\Rightarrow a^2+13\left(7-2a\right)^2-6\left(7-2a\right)-16=0\)
\(\Leftrightarrow53a^2-352a+579=0\Rightarrow\left[{}\begin{matrix}a=\frac{193}{53}>\frac{7}{2}\left(l\right)\\a=3\end{matrix}\right.\) \(\Rightarrow b=1\)
\(\Rightarrow\left\{{}\begin{matrix}2x+3y=9\\5-x-y=1\end{matrix}\right.\) \(\Leftrightarrow...\)