\(C=-\left(x^2+4x+4\right)-\left(y^2-8y+16\right)+22\\ =-\left(x^2+2x.2+2^2\right)-\left(y^2-2.y.4+4^2\right)+22\\ =-\left(x+2\right)^2-\left(y-4\right)^2+22\\ Vậy:max_C=22.khi.x=-2.và.y=4\)

Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

\(D=2023-8x+2y+4xy-y^2-5x^2\)

\(=-\left(y^2+5x^2-4xy-2y+8x-2023\right)\)

\(=-\left(y^2-2.y.\left(2x+1\right)+\left(2x+1\right)^2-\left(2x+1\right)^2+5x^2+8x-2023\right)\)

\(=-\left[\left(y-2x-1\right)^2-4x^2-4x-1+5x^2+8x-2023\right]\)

\(=-\left[\left(y-2x-1\right)^2+x^2+4x-2024\right]\)

\(=-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]+2028\)

Vì \(-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]\le0\forall x,y\)

\(MaxD=2028\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

a)\(A=5-8x-2x^2\)

\(=-2\left(x^2+4x-\frac{5}{2}\right)\)

\(=-2\left(x^2+4x+4-\frac{13}{2}\right)\)

\(=-2\left[\left(x+2\right)^2-\frac{13}{2}\right]\)

\(=-2\left[\left(x+2\right)^2\right]+13\le13\)

Vậy \(A_{max}=13\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

\(-x^2-y^2+xy+2x+2y=-\left[x^2-x\left(y+2\right)+\dfrac{1}{4}\left(y+2\right)^2\right]-\left(\dfrac{3}{4}y^2-3y+3\right)+4=-\left(x-\dfrac{1}{2}y-1\right)^2-\left(\dfrac{\sqrt{3}}{2}y-\sqrt{3}\right)^2+4\le4\)

\(max=4\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

\(A=x^2+y^2-8x-y+68=\left(x-4\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{207}{4}\ge\dfrac{207}{4}\)

\(minA=\dfrac{207}{4}\Leftrightarrow\)\(\left\{{}\begin{matrix}x=4\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(A=x^2-8x+y^2-y+68\)

\(=x^2-8x+16+y^2-y+\dfrac{1}{4}+\dfrac{207}{4}\)

\(=\left(x-4\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{207}{4}\ge\dfrac{207}{4}\forall x,y\)

Dấu '=' xảy ra khi x=4 và \(y=\dfrac{1}{2}\)

\(P=-\left(x^2-4x+4\right)-\left(y^2+4y+4\right)+10\)

\(=-\left(x-2\right)^2-\left(y+2\right)^2+10\le10\)

\(minP=10\Leftrightarrow\) \(\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)