a ) \(C=5x-3x^2+2\)
\(=-3\left(x^2-\dfrac{5}{3}x-\dfrac{2}{3}\right)\)
\(=-3\left(x^2-2x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{49}{36}\right)\)
\(=-3\left[\left(x-\dfrac{5}{6}\right)^2-\dfrac{49}{36}\right]\)
\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{49}{12}\le\dfrac{49}{12}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\)
Vậy GTLN của C là : \(\dfrac{49}{12}\Leftrightarrow x=\dfrac{5}{6}\)
b ) \(D=-8x^2+4xy-y^2+3\)
\(=-\left(4x^2-4xy+y^2\right)-4x^2+3\)
\(=-\left(2x-y\right)^2-4x^2+3\le3\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\4x^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=0\end{matrix}\right.\)
Vậy GTLN của D là : \(3\Leftrightarrow x=y=0\)
