\(D=2x^2+2xy+5y^2-8x-22y\)

\(=2x^2+2x(y-4)+(5y^2-22y)\)

\(=2\left\lbrack x^2+2x\cdot\frac{y-4}{2}+\left(\frac{y-4}{2}\right)^2\right\rbrack-2\left(\frac{y-4}{2}\right)^2+5y^2-22y\)

\(=2\left(x+\frac{y-4}{2}\right)^2-\frac{y^2-8y+16}{2}+5y^2-22y\)

\(=2\left(x+\frac{y-4}{2}\right)^2+\frac92y^2-18y-8\)

\(=2\left(x+\frac{y-4}{2}\right)^2+\frac92\left(y^2-4y+4\right)-26\)

\(=2\left(x+\frac{y-4}{2}\right)^2+\frac92\left(y-2\right)^2-26\ge-26\forall x,y\)

Dấu '=' xảy ra khi y-2=0 và \(x+\frac{y-4}{2}=0\)

=>y=2 và \(x=\frac{-y+4}{2}=\frac{-2+4}{2}=\frac22=1\)

Đặt `A=2x^2+2xy+5y^2-8x-22y`

`<=>2A=4x^2+4xy+10y^2-16x-44y`

`<=>2A=4x^2+4xy+y^2-8(2x+y)+9y^2-28y`

`<=>2A=(2x+y)^2-8(2x+y)+16+9y^2-28y+196/9-196/9`

`<=>2A=(2x+y-4)^2+(3y-14/3)^2-196/9>=-196/9`

`<=>A>=-98/9`

Dấu "=" xảy ra khi `y=14/9,x=(4-y)/2=11/9`

\(F=2x^2+2xy+5y^2-8x-22y\)

<=> \(2F=4x^2+4xy+10y^2-16x-44\)

\(=\left(4x^2+4xy+y^2-16x+16-8y\right)+9y^2-36y-16\)

\(=\left(2x+y-4\right)^2+\left(3y-6\right)^2-52\ge-52\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x+y-4=0\\3y-6=0\end{cases}}\Leftrightarrow y=2;x=1\)

=> min 2F = -52

=> min F = -26. 

Đặt \(A=-2x^2-y^2-2xy+4x+2y+2\)

\(-A=2x^2+y^2+2xy-3x-2y-2\)

\(-A=\left(x^2+2xy+y^2\right)+x^2-4x-2y-2\)

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

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

Mà  \(\left(x+y-1\right)^2\ge0\forall x;y\)

       \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge-4\)

\(\Leftrightarrow A\le4\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x+y-1=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=0\\x=1\end{cases}}\)

Vậy  \(A_{Max}=4\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

Đặt  \(B=x^2-4xy+5y^2+10x-22y+27\)

\(B=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+27\)

\(B=\left[\left(x-2y\right)^2+2\left(x-2y\right)\times5+25\right]+\)\(\left(y^2-2y+1\right)+1\)

\(B=\left(x-2y+5\right)^2+\left(y-1\right)^2+1\)

Mà  \(\left(x-2y+5\right)^2\ge0\forall x;y\)

       \(\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow B\ge1\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy  \(B_{Min}=1\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)

b: \(B=x^2+xy+y^2-3x-3y\)

\(=\frac14\left(4x^2+4xy+4y^2-12x-12y\right)\)

\(=\frac14\left(4x^2+4xy+y^2-12x-6y+3y^2-6y\right)\)

\(=\frac14\left\lbrack\left(2x+y\right)^2-6\left(2x+y\right)+9+3y^2-6y+3-12\right\rbrack\)

\(=\frac14\left\lbrack\left(2x+y-3\right)^2+3\left(y-1\right)^2-12\right\rbrack\ge\frac14\cdot\left(-12\right)=-3\forall x,y\)

Dấu '=' xảy ra khi y-1=0 và 2x+y-3=0

=>y=1 và 2x=-y+3=-1+3=2

=>y=1 và x=1

a: \(A=x^2-2xy+2y^2+2x-10y+17\)

\(=x^2-2xy+y^2+2x-2y+y^2-8y+17\)

\(=\left(x-y\right)^2+2\left(x-y\right)+1+y^2-8y+16\)

\(=\left(x-y+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi y-4=0 và x-y+1=0

=>y=4 và x=y-1=4-1=3

c: \(C=2x^2+9y^2-6xy-6x-12y+2008\)

\(=x^2-6xy+9y^2+4x-12y+x^2-10x+2008\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+4+x^2-10x+25+1979\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1979\ge1979\forall x,y\)

Dấu '=' xảy ra khi x-5=0 và x-3y+2=0

=>x=5 và 3y=x+2=5+2=7

=>x=5 và y=7/3

\(B=2x^2+2xy+5y^2-8x-22y\)

\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+16\right)+\left(4y^2-22y+\frac{484}{16}\right)-\frac{185}{4}\)

\(=\left(x+y\right)^2+\left(x-4\right)^2+\left(2y-\frac{22}{4}\right)^2-\frac{185}{4}\ge-\frac{185}{4}\)

Dấu = xảy ra khi :

.........................

Bn tự giải nốt nhé, mk ko bt có đúng hay ko , nếu sai thì thông cảm nha........