<=> 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/ 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

\(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=3\\ B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\\ B_{min}=51\Leftrightarrow x=5\\ C=\left[\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25\right]+\left(y^2-2y+1\right)+2\\ C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\\ C_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-5=2-5=-3\\y=1\end{matrix}\right.\)

a) \(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)

\(minA=2\Leftrightarrow x=3\)

b) \(B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\)

\(minB=51\Leftrightarrow x=5\)

c) \(C=\left[x^2-2x\left(2y-5\right)+\left(2y-5\right)^2\right]+\left(y^2-2y+1\right)+2=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

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

Bài 6:

a) Ta có: \(A=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu '=' xảy ra khi x=3

b) Ta có: \(B=-x^2-8x+5\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x=-4

c) Ta có: \(C=-x^2+4x+1\)

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

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

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2

Bài 7:

a) Ta có: \(x^2-6x+11\)

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

\(=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=3

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

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

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

\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

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

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

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

`A=x^2-4x+1`
`=x^2-4x+4-3`
`=(x-2)^2-3>=-3`
Dấu "=" xảy ra khi x=2
`B=4x^2+4x+11`
`=4x^2+4x+1+10`
`=(2x+1)^2+10>=10`
Dấu "=" xảy ra khi `x=-1/2`
`C=(x-1)(x+3)(x+2)(x+6)`
`=[(x-1)(x+6)][(x+3)(x+2)]`
`=(x^2+5x-6)(x^2+5x+6)`
`=(x^2+5x)^2-36>=-36`
Dấu "=" xảy ra khi `x=0\or\x=-5`
`D=5-8x-x^2`
`=21-16-8x-x^2`
`=21-(x^2+8x+16)`
`=21-(x+4)^2<=21`
Dấu "=" xảy ra khi `x=-4`
`E=4x-x^2+1`
`=5-4+4-x^2`
`=5-(x^2-4x+4)`
`=5-(x-2)^2<=5`
Dấu "=" xảy ra khi `x=5`

A= x² - 4x +1

   = x² - 4x + 4 - 3

   = (x-2)² -3

Ta có (x-2)² ≥ 0 ∀ x

    ⇒ (x-2)² -3 ≥ -3 ∀ x

Vậy A_Min= -3 tại x=2

B= 4x²+4x+11

  = 4x²+4x+1+10

  = (2x+1)²+10

Ta có (2x+1)² ≥ 0 ∀ x

     ⇒ (2x+1)²+10 ≥ 10 ∀ x

Vậy B_Min=10 tại x= \(\dfrac{-1}{2}\)

C=(x-1)(x+3)(x+2)(x+6)

  = (x-1)(x+6)(x+3)(x+2)

  = (x²+5x-6) (x²+5x+6)

  = (x²+5x)² -36

Ta có (x²+5x)² ≥ 0 ∀ x
  ⇒ (x²+5x)² -36 ≥ -36 ∀ x

Vậy C_Min=-36 tại x=0 hoặc x= -5

\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)

b:

\(D=-25x^2+10x-1-10\)

\(=-\left(25x^2-10x+1\right)-10\)

\(=-\left(5x-1\right)^2-10< =-10\)

Dấu = xảy ra khi x=1/5

\(E=-9x^2-6x-1+20\)

\(=-\left(9x^2+6x+1\right)+20\)

\(=-\left(3x+1\right)^2+20< =20\)

Dấu = xảy ra khi x=-1/3

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

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

Dấu = xảy ra khi x=1