\(2x+3y=5\Rightarrow x=\dfrac{5-3y}{2}\)
\(A=3y^2-2x^2=3y^2-2.\dfrac{\left(5-3y\right)^2}{4}=3y^2-\dfrac{9y^2-30y+25}{2}=\dfrac{6y^2-9y^2+30y-25}{2}=\dfrac{-3y^2+30y-25}{2}=\dfrac{-3\left(y^2-10y+25\right)+50}{2}=\dfrac{-3\left(y-5\right)^2+50}{2}\le\dfrac{50}{2}=25\)Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}y-5=0\\x=\dfrac{5-3y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=5\end{matrix}\right.\)
Vậy \(MaxA=25\)
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