a, \(A=-x^2-2x+3=-\left(x^2+2x-3\right)=-\left(x^2+2x+1-4\right)\)

\(=-\left(x+1\right)^2+4\le4\)

Dấu ''='' xảy ra khi x = -1 

Vậy GTLN là 4 khi x = -1 

b, \(B=-4x^2+4x-3=-\left(4x^2-4x+3\right)=-\left(4x^2-4x+1+2\right)\)

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

Dấu ''='' xảy ra khi x = 1/2 

Vậy GTLN B là -2 khi x = 1/2 

c, \(C=-x^2+6x-15=-\left(x^2-2x+15\right)=-\left(x^2-2x+1+14\right)\)

\(=-\left(x-1\right)^2-14\le-14\)

Vâỵ GTLN C là -14 khi x = 1

Bài 8 : 

b, \(B=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\)

Dấu ''='' xảy ra khi x = 3

Vậy GTNN B là 2 khi x = 3 

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

Dấu ''='' xảy ra khi x = 1/2 

Vậy ...

c, \(x^2-12x+2=x^2-12x+36-34=\left(x-6\right)^2-34\ge-34\)

Dấu ''='' xảy ra khi x = 6

Vậy ...

a) A = x² + 12x + 39

= ( x² + 12x + 36 ) + 3

= ( x + 6 )² + 3 ≥ 3 ∀ x

Đẳng thức xảy ra ⇔ x + 6 = 0 => x = -6

=> MinA = 3 ⇔ x = -6

B = 9x² - 12x 

= 9( x² - 4/3x + 4/9 ) - 4

= 9( x - 2/3 )² - 4 ≥ -4 ∀ x

Đẳng thức xảy ra ⇔ x - 2/3 = 0 => x = 2/3

=> MinB = -4 ⇔ x = 2/3

b) C = 4x - x² + 1

= -( x² - 4x + 4 ) + 5

= -( x - 2 )² + 5 ≤ 5 ∀ x

Đẳng thức xảy ra ⇔ x - 2 = 0 => x = 2

=> MaxC = 5 ⇔ x = 2

D = -4x² + 4x - 3

= -( 4x² - 4x + 1 ) - 2

= -( 2x - 1 )² - 2 ≤ -2 ∀ x

Đẳng thức xảy ra ⇔ 2x - 1 = 0 => x = 1/2

=> MaxD = -2 ⇔ x = 1/2

Ta có A = x² + 12x + 39 = (x² + 12x + 36) + 3 = (x + 6)² + 3 \(\ge\)3

Dấu "=" xảy ra <=> x + 6 = 0

=> x = -6

Vậy Min A = 3 <=> x = -6

Ta có B = 9x² - 12x = [(3x)² - 12x + 4] - 4 =(3x - 2)² - 4 \(\ge\)-4

Dấu "=" xảy ra <=> 3x - 2 =0

=> x = 2/3

Vậy Min B = -4 <=> x = 2/3

b) Ta có C = 4x - x² + 1 = -(x² - 4x - 1) = -(x² - 4x + 4) + 5 = -(x - 2)² + 5 \(\le\)5

Dấu "=" xảy ra <=> x - 2 = 0

=> x = 2

Vậy Max C = 5 <=> x = 2

Ta có D = -4x² + 4x - 3 = -(4x² - 4x + 1) - 2 = -(2x - 1)² - 2 \(\le\)-2

Dấu "=" xảy ra <=> 2x - 1 = 0

=> x = 0,5

Vậy Max D = -2 <=> x = 0,5

Trả lời:

1, A = x³ - 6x² + 12x - 8 = x³ - 3.x².2 + 3.x.2² - 2³ = ( x - 2 )³ 

Thay x = 22 vào A, ta được:

A = ( 22 - 2 )³ = 20³ = 8000

2, B = x² + 4x + 4 = x² + 2.x.2 + 2² = ( x + 2 )²

Thay x = 98 vào B, ta được:

B = ( 98 + 2 )² = 100² = 10000

Ta có A = x³ - 6x² + 12x - 8 = (x - 2)³

Thay x = 22 vào A 

=> A = (22 - 2)³ = 20³ = 8000

b) Sửa đề  B = x² + 4x + 4

= (x + 2)²

Thay x = 98 vào B => B = (98 + 2)²  = 100² = 10000

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a: Ta có: \(A=\left(x-1\right)\left(x-3\right)+11\)

\(=x^2-4x+3+11\)

\(=x^2-4x+4+8\)

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

Dấu '=' xảy ra khi x=2

b: Ta có: \(B=-4x^2+4x+5\)

\(=-\left(4x^2-4x+1-6\right)\)

\(=-\left(2x-1\right)^2+6\le6\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

A = x² - 2x + 9 = ( x² - 2x + 1 ) + 8 = ( x - 1 )² + 8 ≥ 8 ∀ x

Dấu "=" xảy ra khi x = 1

=> MinA = 8 <=> x = 1

B = x² + 6x - 3 = ( x² + 6x + 9 ) - 12 = ( x + 3 )² - 12 ≥ -12 ∀ x

Dấu "=" xảy ra khi x = -3

=> MinB = -12 <=> x = -3

C = ( x - 1 )( x - 3 ) + 9 = x² - 4x + 3 + 9 = ( x² - 4x + 4 ) + 8 = ( x - 2 )² + 8 ≥ 8 ∀ x

Dấu "=" xảy ra khi x = 2

=> MinC = 8 <=> x = 2

D = -x² - 4x + 7 = -( x² + 4x + 4 ) + 11 = -( x + 2 )² + 11 ≤ 11 ∀ x

Dấu "=" xảy ra khi x = -2

=> MaxD = 11 <=> x = -2

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\(B1,a,A=x^2-6x+11\)

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

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

Dấu "=" <=> x=3

Vậy ..........

\(b,B=x^2-20x+101\)

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

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

Dấu "=" <=> x = 10

Vậy .

\(2,a,A=4x-x^2+3\)

            \(=7-\left(x^2-4x+4\right)\)'

             \(=7-\left(x-2\right)^2\le7\)

Dấu ''='' <=> x = 2

Vậy .

\(b,B=-x^2+6x-11\)

       \(=-2-\left(x^2-6x+9\right)\)

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

Dấu ""=" <=> x = 3

Vậy..