\(A=4x-x^2-3=-\left(x^2-4x+3\right)=-\left(x^2-4x+4-1\right)\)

\(A=-\left(\left(x-2\right)^2-1\right)=-\left(x-2\right)^2+1\le1\forall x\)

\(\Rightarrow GTLN\) của A là 1 khi \(-\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

vậy GTLN của A là 1 khi \(x=2\)

\(B=-x^2-4x-2=-\left(x^2+4x+2\right)=-\left(x^2+4x+4-2\right)\)

\(B=-\left(\left(x+2\right)^2-2\right)=-\left(x+2\right)^2+2\le2\forall x\)

\(\Rightarrow GTLN\) của B là 2 khi \(-\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

vậy GTLN của B là 2 khi \(x=-2\)

\(C=2x-2x^2-5=-2\left(x^2-x+\dfrac{5}{2}\right)=-2\left(\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\right)\)

\(C=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\forall x\)

\(\Rightarrow GTLN\) của C là \(-\dfrac{9}{2}\) khi \(-2\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

vậy GTLN của C là \(-\dfrac{9}{2}\) khi \(x=\dfrac{1}{2}\)

\(D=-2x^2-3x+5=-\left(2x^2+3x-5\right)=-\left(\left(\sqrt{2}x+\dfrac{3}{2\sqrt{2}}\right)-\dfrac{49}{8}\right)\)

\(D=-\left(\sqrt{2}x+\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

\(\Rightarrow GTLN\) của D là \(\dfrac{49}{8}\) khi \(-\left(\sqrt{2}x+\dfrac{3}{2\sqrt{2}}\right)=0\Leftrightarrow\sqrt{2}x+\dfrac{3}{2\sqrt{2}}=0\Leftrightarrow\sqrt{2}x=\dfrac{-3}{2\sqrt{2}}\Leftrightarrow x=\dfrac{-3}{4}\)

vậy GTLN của D là \(\dfrac{49}{8}\) khi \(x=\dfrac{-3}{4}\)

A=4x-x²-3

Ta có: \(A=-\left(x^2-4x+3\right)\)

\(=-\left(x^2-2x-2x+3\right)\)

\(=-\left[x\left(x-2\right)-2\left(x-2\right)-1\right]\)

\(=-\left[\left(x-2\right)\left(x-2\right)-1\right]\)

\(=-\left[\left(x-2\right)^2-1\right]\)

Ta có: \(\left(x-2\right)^2-1\ge-1\forall x\Rightarrow-\left[\left(x-2\right)^2-1\right]\le1\forall x\)

Vậy GTLN_A = 1 tại x = 2.

B-x^2-4x-2

Ta có: \(B=x^2-2x-2x-2\)

\(=x\left(x-2\right)-2\left(x-2\right)-6\)

\(=\left(x-2\right)\left(x-2\right)-6\)

\(=\left(x-2\right)^2-6\)

Ta có: \(\left(x-2\right)^2\ge0\forall x\Rightarrow\left(x-2\right)^2-6\ge6\forall x\)

Vậy GTNN_B = 6 tại x = 2.

C=2x-2x^2-5

Ta có: \(C=-2\left(x^2-x+\dfrac{5}{2}\right)\)

\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\) (làm tương tự 2 câu trên)

Ta có: \(-2\left(x-\dfrac{1}{2}\right)^2\le0\forall x\Rightarrow-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\forall x\)

Vậy GTLN_C = \(-\dfrac{9}{2}\) tại x = \(\dfrac{1}{2}\).

D=-2x^2-3x+5

Ta có: \(D=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\) (tương tự câu C)

Ta có: \(-2\left(x+\dfrac{3}{4}\right)^2\le0\forall x\Rightarrow-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Vậy GTLN_D = \(\dfrac{49}{8}\) tại x = \(-\dfrac{3}{4}\).

\(B=4x^2+5y^2-4xy+3x-y\)

\(\Leftrightarrow\left(4x^2-4xy+3x\right)+5y^2-y\)

\(\Leftrightarrow\left[4x^2-4x\left(y-\dfrac{3}{4}\right)+\left(y-\dfrac{3}{4}\right)^2\right]+5y^2-y-y^2+\dfrac{3}{2}y-\dfrac{9}{16}\)\(\Leftrightarrow\left(2x-y+\dfrac{3}{4}\right)^2+\left(4y^2-\dfrac{1}{2}y+\dfrac{1}{64}\right)-\dfrac{37}{64}\)

\(\Leftrightarrow\left(2x-y+\dfrac{3}{4}\right)^2+\left(2y-\dfrac{1}{8}\right)^2-\dfrac{37}{64}\ge\dfrac{-37}{64}\)

Vậy Min B = \(\dfrac{-37}{64}\) khi \(\left[{}\begin{matrix}\left(2x-y+\dfrac{3}{4}\right)^2=0\\\left(2y-\dfrac{1}{8}\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-y+\dfrac{3}{4}=0\\2y-\dfrac{1}{8}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-y+\dfrac{3}{4}=0\\2y=\dfrac{1}{8}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{16}+\dfrac{3}{4}=0\\y=\dfrac{1}{16}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-11}{32}\\y=\dfrac{1}{16}\end{matrix}\right.\)

\(C=9y^2+2x^2-6y-6xy+5x-1\)

\(=\left(9y^2+6y-6xy\right)+2x^2+5x-1\)

\(=\left[9y^2+6y\left(1-x\right)+\left(1-x\right)^2\right]+2x^2+5x-1-1+2x-x^2\)\(=\left(3y-x+1\right)^2+\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{17}{4}\)

\(=\left(3y-x+1\right)^2+\left(x+\dfrac{3}{2}\right)^2-\dfrac{17}{4}\)

Vậy Min C = \(\dfrac{-17}{4}\) khi \(\left[{}\begin{matrix}\left(3y-x+1\right)^2=0\\\left(x+\dfrac{3}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3y-x+1=0\\x+\dfrac{3}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3y-\left(\dfrac{-3}{2}\right)+1=0\\x=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=\dfrac{-5}{6}\\x=\dfrac{-3}{2}\end{matrix}\right.\)