a) \(A=x^2+6x+11\)

\(A=x^2+6x+9+2\)

\(A=\left(x+3\right)^2+2\)

Có: \(\left(x+3\right)^2\ge0\Rightarrow\left(x+3\right)^2+2\ge2\)

Dấu = xảy ra khi: \(\left(x+3\right)^2=0\Rightarrow x+3=0\Rightarrow x=-3\)

Vậy: \(Min_A=2\) tại \(x=-3\)

b) \(B=4x-x^2+1\)

\(B=-x^2+4x-4+5\)

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

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

Có: \(\left(x-2\right)^2\ge0\)

\(\Rightarrow5-\left(x-2\right)^2\le5\)

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

Vậy: \(Max_B=5\) tại \(x=2\)

d, (x-1) (x+2) (x+3) (x+6)
=(x^2+2x-x-2) (x^2+6x+3x+18)
=(x^2-x^2) + (2x-x+6x-3x) = (-2+18)
=0            + (-8x)              =16
=                    x                =16:(-8)
=                  x                  =-2

A = x^2 + 6x + 11

= x^2 + 6x + 9 + 2

= (x + 3)^2 + 2

min = 2

a, A = x^2 + 6x + 11

= x^2 + 6x + 9 + 2

= (x + 3)^2 + 2

làm tiếp

b, x^2 - 20x + 101

= x^2  20x + 100 + 1

= (x - 10)^2 + 1

có (x - 10)^2 > 0 => (x - 10)^2 +  > 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

<=> xaa ) C= x²-6x + 11= (x-3)² +2

ta co : (x-3)² + > hoặc = 2

=> C đạt giá trị nhỏ nhất khi C=2

<=> x=3

b) D =(x-1) (x+2)(x+3)(x+6)

= [ (x-1)(x+6)][(x+2)(x+3)]

=(x² +5x -6)(x²+5x +6)

=(x²+5x )² - 36

ta có (x² +5x)² -36 luôn > hoặc = -36

=> D đạt GTNN khi D = -36

<=>(x² + 5x)^{2 =0}

=> x = 0 hoac x =-5

c) E = x² - 4x + y² - 8y + 6

=(x² -4x +4 ) + (y² - 8y +16 ) -14

= (x -2)² +( y-4)² -14

ta co (x-2)² + (y-4)² -14 luôn > hoặc = -14

=> E dat GTNN khi E = -14

<=> (x-2)²​ =0 va (y-4)² =0

<=> x =2 va y=4

d) G =x² -4xy +5y² + 10x -22y + 28 ( de sai nha ban )

= [(x² - 4xy + 4y² ) + 10x -20y +25 ]+ ( y² -2y +1 ) +2

= [(x-2y)² + 10x - 20y + 25 ] + (y-1)² +2

= [( x-2y)² + 2. 5 (x-2y) + 25 ] + (y-1)² +2

= (x-2y +5)² + ( y-1)² +2

ta co (x-2y +5 )² + (y-1)² +2 luôn > hoặc = 0

=> G đạt GTNN khi (x-2y+5 )²=0 hoac (y-1)² =0

<=> y-1 = 0 => y = 1

,=> x =-3

\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow x=-2\)

\(B=x^2+10x-1=\left(x+5\right)^2-26\ge-26\)

Dấu \("="\Leftrightarrow x=-5\)

\(C=5-4x+4x^2=\left(2x-1\right)^2+4\ge4\)

Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)

\(D=x^2+y^2-2x+6y-3=\left(x-1\right)^2+\left(y+3\right)^2-13\ge-13\)

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

\(E=2x^2+y^2+2xy+2x+3=\left(x+y\right)^2+\left(x+1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow x=-y=-1\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

\(A=x^2+4x+5\)

\(=x^2+4x+4+1\)

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

Dấu '=' xảy ra khi x=-2

\(C=4x^2-4x+5\)

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

\(=\left(2x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a/ Ta có:

\(A=x^2-6x+11\)

\(A=x\cdot x-3x-3x+3\cdot3+2\)

\(A=x\left(x-3\right)-3\left(x-3\right)+2\)

\(A=\left(x-3\right)\left(x-3\right)+2\)

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

Vì \(\left(x-3\right)^2\ge0\)

Nên GTNN của \(\left(x-3\right)^2\)là 0

=> \(A_{min}=0+2=2\)

mình chỉ biết a. thôi

a) ta có : \(A=x^2-6x+11\)

\(A=x.x-3x-3x+3.3+2\)

\(A=x\left(x-3\right)-3\left(x-3\right)+2\)

\(A=\left(x-3\right)\left(x-3\right)+2\)

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

vì \(\left(x-3\right)^2\ge0\)

nên GTNN của \(\left(x-3\right)^2\)là \(0\)

\(\Rightarrow\)\(A_{min}\)\(=0+2=2\)

oOo Không đủ can đảm để oOo copy mà nói nhưu mk tự làm

\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(A=x^2-10x+3=\left(x^2-10x+25\right)-22=\left(x-5\right)^2-22\ge-22\)

Vậy GTNN của A là -22 khi x = 5

\(B=x^2+6x-5=\left(x^2+6x+9\right)-14=\left(x+3\right)^2-14\ge-14\)

Vậy GTNN của B là -14 khi x = -3

\(C=x\left(x-3\right)=x^2-3x=\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)

Vậy GTNN của C là \(-\dfrac{9}{4}\) khi x = \(\dfrac{3}{2}\)

\(D=x^2+y^2-4x+20=\left(x^2-4x+4\right)+y^2+16=\left(x-2\right)^2+y^2+16\ge16\)

Vậy GTNN của D là 16 khi x = 2; y = 0

\(E=x^2+2y^2-2xy+4x-6y+100\)

\(E=\left(x^2+y^2+4-2xy+4x-4y\right)+\left(y^2-2y+1\right)+95\)

\(E=\left(x-y+2\right)^2+\left(y-1\right)^2+95\ge95\)

Vậy GTNN của E là 95 khi x = -1 ; y = 1

\(F=2x^2+y^2-2xy+4x+100\)

\(F=\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)+96\)

\(F=\left(x-y\right)^2+\left(x+2\right)^2+96\ge96\)

Vậy GTNN của F là 96 khi x = -2; y = -2

\(A=-x^2-12x+3=-\left(x^2+12x+36\right)+39=-\left(x+6\right)^2+39\le39\)

Vậy GTLN của A là 39 khi x = -6

\(B=7-4x^2+4x=-\left(4x^2-4x+1\right)+8=-\left(2x-1\right)^2+8\le8\)

Vậy GTLN của B là 8 khi x = \(\dfrac{1}{2}\)