a)  ... = (x^2 -2xy + y^2)+(x^2 -2x+1)+2014=(x-y)^2 + (x-1)^2 +2014 >= 2014 

Đăngt thức xay ra khi x=y=1

\(B=2x^2+y^2-8x+2xy-4y+2025\)

\(=x^2+2xy+y^2-4x-4y+x^2-4x+2025\)

\(=\left(x+y\right)^2-4\left(x+y\right)+4+x^2-4x+4+2017\)

\(=\left(x+y-2\right)^2+\left(x-2\right)^2+2017\ge2017\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-2=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=-x+2=-2+2=0\end{cases}\)

Sửa đề: \(D=4x^2+y^2+2xy-8x+2y+5\)

\(=x^2+2xy+y^2+2x+2y+3x^2-10x+5\)

\(=\left(x+y\right)^2+2\left(x+y\right)+1+3x^2-10x+4\)

\(=\left(x+y+1\right)^2+3\left(x^2-\frac{10}{3}x+\frac43\right)\)

\(=\left(x+y+1\right)^2+3\left(x^2-2\cdot x\cdot\frac53+\frac{25}{9}-\frac{13}{9}\right)=\left(x+y+1\right)^2+3\left(x-\frac53\right)^2-\frac{13}{3}\ge-\frac{13}{3}\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x+y+1=0\\ x-\frac53=0\end{cases}\Rightarrow\begin{cases}x=\frac53\\ y=-x-1=-\frac53-1=-\frac83\end{cases}\)

B=(x²+2,x,y+y²)+(x²-2.x.4+4²)+2012

B=(x+y)²+(x-4)²+2012

   (x+y)² lớn hoăc bằng 0 (mình ko ghi dc ki hiệu)

  (x-4)² lớn hoăc bằng 0 (mình ko ghi dc ki hiệu)

=>(x+y)²+(x-4)²+2012 lớn hoăc bằng 2012

Dấu = xảy ra khi x+y=0 => x=-4

                          x-4=0 => x=4

Giải PT: \(x^2+3y^2+2xy-8x-16y+23=0\)

\(\Leftrightarrow x^2+y^2+16+2xy-8x-8y+2y^2-8y+7=0\)

\(\Leftrightarrow\left(x+y-4\right)^2+2\left(y^2-4y+4\right)-1=0\)

\(\Leftrightarrow\left(x+y-4\right)^2+2\left(y-2\right)^2-1=0\)

\(\Rightarrow\left(x+y-4\right)^2=-2\left(y-2\right)^2+1\le1\)

Dấu "=" xảy ra khi : \(-2\left(y-2\right)^2=0\Rightarrow y=2\)

\(\Rightarrow\)\(\text{│}x+y-4\text{│}\le1\)

\(\Rightarrow-1\le x+y-4\le1\)

\(\Rightarrow3\le x+y\le5\)

Vậy B_min=3 khi y=2;x=1

       B_max=5 khi y=2;x=3

\(P=3x^2+y^2-8x+2xy+16\)

\(P=\left(x^2+2xy+y^2\right)+\left(2x^2-8x+8\right)+8\)

\(P=\left(x+y\right)^2+2\left(x-2\right)^2+8\ge8\)

Vậy GTNN của P=8 <=> \(\orbr{\begin{cases}x+y=0\\x-2=0\end{cases}}\)<=>\(\orbr{\begin{cases}y=-2\\x=2\end{cases}}\)

\(A=2x^2+y^2+2xy+60+8x+8y\)

    \(=\left(x^2+y^2+2xy\right)+8x+8y+16+y^2+44\)

    \(=\left(x+y\right)^2+2\left(x+y\right).4+16+y^2+44\)

    \(=\left(x+y+4\right)^2+y^2+44\)

Vì \(\hept{\begin{cases}\left(x+y+4\right)^2\ge0\forall x\\y^2\ge0\forall y\end{cases}}\)

\(\Rightarrow A\ge44\)

Dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y+4=0\\y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-4\\y=0\end{cases}}\)

Vậy \(minA=44\Leftrightarrow\hept{\begin{cases}x=-4\\y=0\end{cases}}\)

bạn cũng dc đó

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`

m ra gtnn là -26