= x^2-4xy+4y^2+y^2-22y+121-93

=(x+2y)^2+(y-11)^2>=-93

GNNN là -93

Ta có: \(B=x^2-4xy+5y^2-22y+28\)

                \(=x^2-4xy+y^2-22y+121-93\)

                  \(=\left(x-2y\right)^2+\left(y-11\right)^2-93\)

Vì \(\left(x-2y\right)^2\ge0;\left(y-11\right)^2\ge0\)

\(\Rightarrow B\ge-93\)

Dấu "=" xảy ra khi \(y-11=0\Rightarrow y=11\)

                              \(x-2y=0\Rightarrow x-2.11=0\Rightarrow x=22\)

Vậy B_min=-93 khi x=22; y=11

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

\(=x^2-2xy+y^2+4y^2+4y+1+2\)

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

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

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(2y+1\right)^2\ge0\forall y\end{matrix}\right.\)

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

\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{1}{2}\)

Vậy \(D_{min}=2\Leftrightarrow x=y=-\dfrac{1}{2}\)

d= x²  + 5y² + 2xy - 2y + 2005

d= x² + 2xy + y²  + 4y²  - 2y + \(\frac{1}{4}+\)

d= ( x+ y )²  + ( 2y - \(\frac{1}{2}\))²  + \(\frac{8019}{4}\)\(\ge\)\(\frac{8019}{4}\)

dmin= \(\frac{8019}{4}khi\hept{\begin{cases}y=\frac{1}{4}\\x=-y=\frac{-1}{4}\end{cases}}\)

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. 

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

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

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

Dấu = xảy ra khi :

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