a) \(A=6x-x^2-11=-\left(x^2-6x+9\right)-2=-\left(x-3\right)^2-2\le-2\)

Dấu \(=\)khi \(x-3=0\Leftrightarrow x=3\).

b) \(B=x^2-5x-2=x^2-2.\frac{5}{2}x+\left(\frac{5}{2}\right)^2-\frac{33}{4}=\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\ge-\frac{33}{44}\)

Dấu \(=\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\).

a) Đặt \(A=10+2x-5x^2\)

\(-A=5x^2-2x-10\)

\(-5A=25x^2-10x-50\)

\(-5A=\left(25x^2-10x+1\right)-51\)

\(-5A=\left(5x-1\right)^2-51\)

Do \(\left(5x-1\right)^2\ge0\forall x\)

\(\Rightarrow-5A\ge-51\)

\(A\le\frac{51}{5}\)

Dấu "=" xảy ra khi : \(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

Vậy Max A = \(\frac{51}{5}\Leftrightarrow x=\frac{1}{5}\)

b) Đặt \(B=x^2-6x+10\)

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

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

Mà \(\left(x-3\right)^2\ge0\forall x\)

\(B\ge1\)

Dấu "=" xảy ra khi :

\(x-3=0\Leftrightarrow x=3\)

Vậy Min B \(=1\Leftrightarrow x=3\)

A = x² + 5x + 7 

   = ( x² + 5x + 25/4 ) + 3/4

   = ( x + 5/2 )² + 3/4

\(\left(x+\frac{5}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Đẳng thức xảy ra <=> x + 5/2 = 0 => x = -5/2

=> MinA = 3/4 <=> x = -5/2

B = 6x - x² - 5

   = -( x² - 6x + 9 ) + 4

   = -( x - 3 )² + 4

\(-\left(x-3\right)^2\le0\forall x\Rightarrow-\left(x-3\right)^2+4\le4\)

Đẳng thức xảy ra <=> x - 3 = 0 => x = 3

=> MaxB = 4 <=> x = 3

C = ( 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

\(\left(x^2+5x\right)^2\ge0\forall x\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)

Đẳng thức xảy ra <=> x² + 5x = 0

                             <=> x( x + 5 ) = 0

                             <=> x = 0 hoặc x = -5

=> MinC = -36 <=> x = 0 hoặc x = -5

Thank bn.😊😉

1/

a, \(A=4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\ge4\)

Dấu "=" xảy ra khi x=1/2

Vậy Amin=4 khi x=1/2

b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)

Dấu "=" xảy ra khi x=-1

Vậy Bmin = -4 khi x=-1

2/

a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)

Dấu "=" xảy ra khi x=3

Vậy Amax = 19 khi x=3

b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmax = 31/8 khi x=5/4

\(A=\left(x+3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=-3\\ B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{29}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{29}{4}\ge-\dfrac{29}{4}\\ B_{min}=-\dfrac{29}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ C=\left(9x^2-12x+4\right)+2017=\left(3x-2\right)^2+2017\ge2017\\ C_{min}=2017\Leftrightarrow x=\dfrac{2}{3}\)

toi ko bt

A= -4 - x^2 +6x

  =-(x²-6x+9)+5

=-(x-3)²+5\(\le\)5

Dấu "=" xảy ra khi x=3

Vậy...............

B= 3x^2 -5x +7

\(=3\left(x^2-2.\frac{5}{6}x+\frac{25}{36}\right)-\frac{59}{12}\)

\(=3\left(x-\frac{5}{6}\right)^2-\frac{59}{12}\ge\frac{-59}{12}\)

Dấu "=" xảy ra khi \(x=\frac{5}{6}\)

Vậy.................

A= 5x-x²= -x²+5x = -(x²-5x+25/4-25/4)= -(x-5/2)²+25/4

vì -(x-5/2)²< hoặc = 0 vs mọi x

nên - (x-5/2)+25/4< hoặc =25/4

dấu bằng xảy ra khi và chỉ khi x-5/2=0

=> x=5/2

câu b tg tự đặt dấu trừ ra ngoài rồi tách 11= 9+2 là ra giá trị lớn nhất của B=-2 tại x=3

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)