\(A=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\le10\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy GTLN của A là : \(10\Leftrightarrow x=3\)
\(B=-x^2-4x-2\)
\(=-\left(x^2+4x+2\right)\)
\(=-\left(x^2+4x+4-2\right)\)
\(=-\left[\left(x+2\right)^2-2\right]\)
\(=-\left(x+2\right)^2+2\le2\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy GTLN của B là : \(2\Leftrightarrow x=-2\)
C ) Sai đề
\(D=\left(2-x\right)\left(3x+4\right)\)
\(=6x-3x^2+8-4x\)
\(=-3x^2+2x+8\)
\(=-3\left(x^2-\dfrac{2}{3}x-\dfrac{8}{3}\right)\)
\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}-\dfrac{25}{9}\right)\)
\(=-3\left[\left(x-\dfrac{1}{3}\right)^2-\dfrac{25}{9}\right]\)
\(=-3\left(x-\dfrac{1}{3}\right)^2+\dfrac{25}{3}\le\dfrac{25}{3}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy GTLN của D là : \(\dfrac{25}{3}\Leftrightarrow x=\dfrac{1}{3}\)
\(E=-8x^2+4xy-y^2+3\)
\(=-8x^2+4xy-\dfrac{y^2}{2}-\dfrac{y^2}{2}+3\)
\(=-2\left[4x^2-2xy+\dfrac{y^2}{4}\right]-\dfrac{y^2}{2}+3\)
\(=-2\left(2x-\dfrac{y}{2}\right)^2-\dfrac{y^2}{2}+3\le3\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-\dfrac{y}{2}\right)^2=0\\\dfrac{y^2}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-\dfrac{y}{2}=0\\y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{y}{2}\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=0\\y=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy GTLN của E là : \(3\Leftrightarrow x=y=0\)
