a, (x-1)(x-3)+11

=x²-3x-x+3+11

=(x-2)²+10

Vì..................................

b,5-4x²+4x

=-(4x²-4x+4)+9

=-(2x-2)²+9

...........................................................

1:

a: =x^2-7x+49/4-5/4

=(x-7/2)^2-5/4>=-5/4

Dấu = xảy ra khi x=7/2

b: =x^2+x+1/4-13/4

=(x+1/2)^2-13/4>=-13/4

Dấu = xảy ra khi x=-1/2

e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

f: x^2-4x+7

=x^2-4x+4+3

=(x-2)^2+3>=3

Dấu = xảy ra khi x=2

2:

a: A=2x^2+4x+9

=2x^2+4x+2+7

=2(x^2+2x+1)+7

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: x^2+2x+4

=x^2+2x+1+3

=(x+1)^2+3>=3

Dấu = xảy ra khi x=-1

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3

đơn giản wá 

a) \(A=x^2-3x-x+3+11\) 

      \(=\left(x^2-4x+4\right)+10\)

      \(=\left(x-2\right)^2+10\ge10\forall x\in R\) 

Dấu "=" xảy ra<=> \(\left(x-2\right)^2=0\Leftrightarrow x=2\) 

b) \(B=5-4x^2+4x\) 

      \(=-\left(4x^2-4x+1\right)+6\) 

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

Dấu "=" xảy ra<=> \(-\left(2x-1\right)^2=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

       \(=\left(x^2-3x\right)^2-1\ge-1\forall x\in R\)

Dấu "=" xảy ra<=>\(\left(x^2-3x\right)^2=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow x=0;x=3\) 

I zì:vv

a) Ta có: \(A=4x^2+4x+11=4x^2+4x+1=10=\left(2x+1\right)^2+10\ge10\forall x\)

Vậy Min_A=10 khi \(x=-\dfrac{1}{2}\)

b) Ta có: \(B=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)\)

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

Vậy Max_B=21 khi x=-4

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

suy ra Amin=-1

\(B=4x^2+4x+11=4\left(x^2+x+\frac{11}{4}\right)=4\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{10}{4}\right)=4\left(x+\frac{1}{2}\right)^2+10\) Suy ra Bmin = 10

phần B có bạn làm rồi nha mình không làm nữa

A=x²-4x+1=x²-4x+4-3=(x-2)²-3

Vì (x-2)²\(\ge\)0\(\forall\)x \(\Rightarrow\)(x-2)²-3\(\ge\)-3\(\forall\)x

Vậy minA = -3

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

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

C=(x²+5x-6)(x²+5x+6)

Đặt x²+5x+6=t . Ta có:

C= (t-12).t=t²-12t=t²-12+36-36=(t-6)²-36

C= (x²+5x+6-6)²-36=(x²+5x)²-36

Vì (x²+5x)²\(\ge\)0\(\forall\)x \(\Rightarrow\)(x²+5x)²-36\(\ge\)-36\(\forall\)x

Vậy minC= -36

D=5-8x-x²=-(x²+8x-5)=-(x²+8x+16-21)=-\(\left[\left(x+4\right)^2-21\right]\)

D=-(x+4)²+21=21-(x+4)²

Vì (x+4)²\(\ge\)0\(\forall\)x\(\Rightarrow\)21-(x+4)²\(\le\)21\(\forall\)x

Vậy maxD=21

E=4x-x²+1=-(x²-4x-1)=-(x²-4x+4-5)=-\(\left[\left(x-2\right)^2-5\right]\)=-(x-2)²+5=5-(x-2)²

Vì (x-2)²\(\ge0\forall x\)\(\Rightarrow\)5-(x-2)²\(\le5\forall x\)

Vậy maxE=5

a.

\(A=\dfrac{2013}{x^2}-\dfrac{2}{x}+1=2013\left(\dfrac{1}{x}-\dfrac{1}{2013}\right)^2+\dfrac{2012}{2013}\ge\dfrac{2012}{2013}\)

Dấu "=" xảy ra khi \(x=2013\)

b.

\(B=\dfrac{4x^2+2-4x^2+4x-1}{4x^2+2}=1-\dfrac{\left(2x-1\right)^2}{4x^2+2}\le1\)

\(B_{max}=1\) khi \(x=\dfrac{1}{2}\)

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

\(B_{max}=-\dfrac{1}{2}\) khi \(x=-1\)

a: =-x^2+6x-4

=-(x^2-6x+4)

=-(x^2-6x+9-5)

=-(x-3)^2+5<=5

Dấu = xảy ra khi x=3

b: =3(x^2-5/3x+7/3)

=3(x^2-2*x*5/6+25/36+59/36)

=3(x-5/6)^2+59/12>=59/12

Dấu = xảy ra khi x=5/6

c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)

\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)

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

Dấu = xảy ra khi x=4 hoặc x=2