2:

a: =-(x^2-12x-20)

=-(x^2-12x+36-56)

=-(x-6)^2+56<=56

Dấu = xảy ra khi x=6

b: =-(x^2+6x-7)

=-(x^2+6x+9-16)

=-(x+3)^2+16<=16

Dấu = xảy ra khi x=-3

c: =-(x^2-x-1)

=-(x^2-x+1/4-5/4)

=-(x-1/2)^2+5/4<=5/4

Dấu = xảy ra khi x=1/2

1) 

a) \(A=x^2+4x+17\)

\(A=x^2+4x+4+13\)

\(A=\left(x+2\right)^2+13\) 

Mà: \(\left(x+2\right)^2\ge0\) nên \(A=\left(x+2\right)^2+13\ge13\)

Dấu "=" xảy ra: \(\left(x+2\right)^2+13=13\Leftrightarrow x=-2\)

Vậy: \(A_{min}=13\) khi \(x=-2\)

b) \(B=x^2-8x+100\)

\(B=x^2-8x+16+84\)

\(B=\left(x-4\right)^2+84\)

Mà: \(\left(x-4\right)^2\ge0\) nên: \(A=\left(x-4\right)^2+84\ge84\)

Dấu "=" xảy ra: \(\left(x-4\right)^2+84=84\Leftrightarrow x=4\)

Vậy: \(B_{min}=84\) khi \(x=4\)

c) \(C=x^2+x+5\)

\(C=x^2+x+\dfrac{1}{4}+\dfrac{19}{4}\)

\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\)

Mà: \(\left(x+\dfrac{1}{2}\right)^2\ge0\) nên \(A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Dấu "=" xảy ra: \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=\dfrac{19}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy: \(A_{min}=\dfrac{19}{4}\) khi \(x=-\dfrac{1}{2}\)

1:

a: A=x^2+4x+4+13

=(x+2)^2+13>=13

Dấu = xảy ra khi x=-2

b; =x^2-8x+16+84

=(x-4)^2+84>=84

Dấu = xảy ra khi x=4

c: =x^2+x+1/4+19/4

=(x+1/2)^2+19/4>=19/4

Dấu = xảy ra khi x=-1/2

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2

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