\(\left\{{}\begin{matrix}2x+y=5\\x-2y=10a+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+2y=10\\x-2y=10a+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=10a+15\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2a+3\\y=-4a-1\end{matrix}\right.\)
Ta có \(xy=\left(2a+3\right)\left(-4a-1\right)=-8a^2-14a-3=-8\left(a^2+\frac{14}{8}a+\frac{3}{8}\right)\)
\(=-8\left[\left(a+\frac{7}{8}\right)^2-\frac{25}{64}\right]=\frac{25}{8}-8\left(a+\frac{7}{8}\right)^2\le\frac{25}{8}\)
Vậy GTLN của \(xy=\frac{25}{8}\) khi và chỉ khi \(\left(a+\frac{7}{8}\right)^2=0\Leftrightarrow a=-\frac{7}{8}\)
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