Ta có: 5x2+10y2-6xy-4x-2y +3= x2 -6xy +(3y)2 +4x2 +y2 -4x -2y +3

= (x - 3y)2 +(2x)2 -4x+1+ y2 -2y+1 +1

= (x-3y)2 + (2x -1)2 + (y-1)2 +1

Ta có :(x-3y)2 luôn lớn hơn hoặc bằng 0

(2x -1)2 luôn lớn hơn hoặc bằng 0

(y-1)2 luôn lớn hơn hoặc bằng 0

=>(x-3y)2 + (2x -1)2 + (y-1)2 luôn lớn hơn hoặc bằng 0

=>(x-3y)2 + (2x -1)2 + (y-1)2 +1 >0

ta có:\(A=x^2+5y^2-4xy-2y+2x+2010\)

\(=x^2+4y^2+y^2-4xy-4y+2y+2x+1+1+2008\)

\(=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1+\left(y^2+2x+1\right)+2008\)

\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y+1\right)^2+2008\)

\(=\left(x-2y+1\right)^2+\left(y+1\right)^2+2008\)

Vì: (x-2y+1)²+(y+1)>0 với \(\forall x;y\)

do đó: (x-2y+1)²+(y+1)+2008 > 2008 với \(\forall x;y\)

Dấu "=" xảy ra khi x-2y+1=0 và y+1=0

ta có:

y+1=0=>y=0-1=>y=-1

thay y=-1 và x-2y+1=0

=>x-2.(-1)+1=0

=>x+2+1=0

=>x+2=-1

=>x=-1-2

=>x=-3

vậy \(A_{min}=2008\) khi x=-3 hoặc x=-1

đặt biểu thức là A. Ta có:

A=x² - 4xy + 5y² - 2y + 28

  = (x²-4xy+4y²) + (y²-2y +1)+27

  =(x-2y)² + (y-1)² + 27

vì (x-2y)² ≥ 0; (y-1)² ≥ 0 ⇔ A ≥ 27

⇔\(\left[\begin{array}{} (x-2y)^2=0\\ (y-1)^2 =0 \end{array} \right.\)           ⇔\(\left[\begin{array}{} x=2\\ y=1\end{array} \right.\)

Vậy, Min A=27 khi x=2; y=1

F = 5x² + 2y² + 4xy - 2x + 4y + 8

F = ( 4x² + 4xy + y² ) + ( x² - 2x + 1 ) + ( y² + 4y + 4 ) + 3

F = ( 2x + y )² + ( x - 1 )² + ( y + 2 )² + 3

\(\hept{\begin{cases}\left(2x+y\right)^2\\\left(x-1\right)^2\\\left(y+2\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\forall x,y\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}2x+y=0\\x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

Vậy MinF = 3 <=> x = 1 , y = -2

G = 5x² + 5y² + 8xy + 2y + 2020

= x² + ( 4x² + 8xy + 4y² ) + ( y² + 2y + 1 ) + 2019

= x² + ( 2x + 2y )² + ( y + 1 )² + 2019

\(\hept{\begin{cases}x^2\\\left(2x+2y\right)^2\\\left(y+1\right)^2\end{cases}}\ge0\forall x,y\Rightarrow x^2+\left(2x+2y\right)^2+\left(y+1\right)^2+2019\ge2019\forall x,y\)

Tuy nhiên đẳng thức không xảy ra :P

\(VT=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\left(x-y\right)^2\left(x+y\right)^2=VP\)

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

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

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

\(-x^2+4xy-5y^2-8y-18\)

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

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

Vì \(-\left(x+2y\right)^2\le0;-\left(y+4\right)^2\le\forall x;y\)

\(\Rightarrow-\left(x+2y\right)^2-\left(y+4\right)^2-2< 0\forall x;y\)

\(\Rightarrow dpcm\)

a) \(-x^2+4xy-5y^2-8y-18=-\left(x^2-4xy+5y^2+8y+18\right)\)

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

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

Vì \(\left(x-2y\right)^2\ge0\forall x,y\); \(\left(y+4\right)^2\ge0\forall y\); \(2>0\)

\(\Rightarrow\left(x-2y\right)^2+\left(y+4\right)^2+2>0\)

\(\Rightarrow-\left[\left(x-2y\right)^2+\left(y+4\right)^2+2\right]< 0\)

\(\Rightarrow-x^2+4xy-5y^2-8y-18\)luôn âm với mọi x ( đpcm )

ta có D=x^2 +2.y^2 -2xy+4x-5y-12

<=>D=(x^2 +y^2 +4 -2xy-4y+4x) +[y^2 -2.y.(1/2) +1/4] -1/4+8

<=>D=(x-y+2)^2 +(y-1/2)^2  +31/4

mà (x-y+2)^2 >= 0 và (y-1/2)^2>=0 nên (x-y+2)^2 +(y-1/2)^2 +31/4 >= 31/4

dấu '=' xảy ra khi :y-1/2=0 và x-y+2=0 <=> y=1/2 và x=-3/2

vậy GTNN của D là 31/4 khi x=-3/2, y=1/2

\(A=x^2-20x+101=\left(x-10\right)^2+1\ge1\)

\(minA=1\Leftrightarrow x=10\)

\(B=2x^2+40x-1=2\left(x+10\right)^2-201\ge-201\)

\(minB=-201\Leftrightarrow x=-10\)

\(C=x^2-4xy+5y^2-2y+28=\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+27=\left(x-2y\right)^2+\left(y-1\right)^2+27\ge27\)

\(minC=27\Leftrightarrow\)\(\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

\(D=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x+10\right)=\left(x^2-7x\right)^2-100\ge-100\)

\(minD=100\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

a: Ta có: \(A=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=10

b: ta có: \(B=2x^2+40x-1\)

\(=2\left(x^2+20x-\dfrac{1}{2}\right)\)

\(=2\left(x^2+20x+100-\dfrac{201}{2}\right)\)

\(=2\left(x+10\right)^2-201\ge-201\forall x\)

Dấu '=' xảy ra khi x=-10

a,   B=x²+4xy+y²+x²-8x+16+2012

       B=(x+y) ²+(x-4)²+2012

 Vậy B >=2012 ( Dấu "=" xảy ra khi x=4,y=-4)

b làm tương tự 

c,  9x²+6x+1+y²-4y+4+x²-4xz+4z²=0

     (3x+1)²+(y-4)²+(x-2z)²=0

    Vậy 3x+1=0 => x = -1/3

           y-4=0 => y=4

             x-2z=0  thế x=-1/3 ta được.      -1/3-2z=0 => z = -1/6

Bạn nhớ ghi lại đề minh không ghi đề 

a) \(B=2x^2+y^2+2xy-8x+2028\)

\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+4^2\right)+2012=\left(x+y\right)^2+\left(x-4\right)^2+2012\ge2012\)

\(MinB=2012\Leftrightarrow\hept{\begin{cases}x=4\\y=-4\end{cases}}\)

b)\(C=x^2+5y^2+4xy+2x+2y-7\)

\(=\left(x^2+4xy+4y^2\right)+\left(2x+4y\right)+1+\left(y^2-2y+1\right)-9\)

\(=\left(\left(x+2y\right)^2+2\left(x+2y\right)+1\right)+\left(y-1\right)^2-9=\left(x+2y+1\right)^2+\left(y-1\right)^2-9\ge9\)

\(MinC=-9\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

c)\(10x^2+y^2+4z^2+6x-4y-4xz+5=0\)

\(\Leftrightarrow\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)^2+\left(y-2\right)^2+\left(x-2z\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}\)