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

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

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

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

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

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

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

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

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)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

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

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

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

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

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

B=[(x - 2)(x - 5)](x²– 7x - 10) 
= (x²- 7x + 10)(x² - 7x - 10)
= (x² - 7x)²- 10²
= (x² - 7x)² - 100

=>(x²-7x)²\(\ge\) 100

GTNN = -100 \(\Rightarrow\) x² - 7x = 0 \(\Leftrightarrow\) x(x-7) = 0 \(\Leftrightarrow\) x = 0 hoặc x = 7

B = x² - 4xy + 5y² + 10x - 22y + 28 
= x² - 4xy + 4y²+ y²+ 10(x-2y) + 28 
= (x - 2y)²+ 10(x-2y) + 25 + y²- 2y+ 1 + 2 
= (x-2y + 5)² + (y-1)² + 2\(\ge\) 2 
GTNN B = 2, khi y=1, x=-3

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

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

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

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: 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

\(A=x^2+5y^2-4xy-2y+2x+2010\)

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

\(=\left(x-2y+1\right)^2+\left(y+1\right)^2+2008\ge2008\)

\(minA=2008\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)

\(A=\left[\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+1\right]+\left(y^2+2y+1\right)+2008\\ A=\left[\left(x-2y\right)^2+2\left(x-2y\right)+1\right]+\left(y+1\right)^2+2008\\ A=\left(x-2y+1\right)^2+\left(y+1\right)^2+2008\ge2008\\ A_{min}=2008\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)

Đáp án D

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