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 ...

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

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

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

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..

cau A thay = bằng cộng ạ

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

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

a chưa bt làm

b) 

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

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

\(D=-\left(x^2-2\cdot x\cdot3+3^2+2\right)\)

\(D=-\left[\left(x-3\right)^2+2\right]\)

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

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

mà ( x - 3 )² >= 0 với mọi x

\(\Rightarrow D\le-2\)

Dấu "=" xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy Dmax = -2 <=> x = 3

thanks b

a) \(C=4x-x^2+3=-\left(x^2-4x-3\right)\)

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

\(=-\left[\left(x-2\right)^2-7\right]\)

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

Vậy gt lớn nhất của bt là 7\(\Leftrightarrow x=2\)

a: \(A=\dfrac{x^4+x^2+11x^2+11}{x^4+x^2+5x^2+5}=\dfrac{\left(x^2+11\right)\left(x^2+1\right)}{\left(x^2+5\right)\left(x^2+1\right)}=\dfrac{x^2+11}{x^2+5}\)

b: \(A=\dfrac{x^2+5+6}{x^2+5}=1+\dfrac{6}{x^2+5}< =1+\dfrac{6}{5}=\dfrac{11}{5}\)

Dấu = xảy ra khi x=0

\(-2x^2+6x-11=\left(-2\right)\left(x^2-3x+\frac{11}{2}\right)\)

\(=\left(-2\right)\left[x^2-2.\frac{3}{2}x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+\frac{11}{2}\right]\)

\(=\left(-2\right)\left[\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\right]\)

\(=\left(-2\right)\left(x-\frac{3}{2}\right)^2-\frac{13}{2}\le-\frac{13}{2}\)

Vậy Max = -13/2 khi x - 3/2 = 0 => x = 3/2