1.

$x(x+2)(x+4)(x+6)+8$

$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$

$=a(a+8)+8$ (đặt $x^2+6x=a$)

$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$

Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$

2.

$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$

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

$=5-[(x^2+5x)^2-6^2]$

$=41-(x^2+5x)^2\leq 41$

Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$

3.

Đặt $x+3=a; 7-x=b$ thì $a+b=10$ 

$C=a^4+b^4$

Áp dụng BĐT Bunhiacopxky:

$(a^4+b^4)(1+1)\geq (a^2+b^2)^2$

$\Rightarrow C\geq \frac{(a^2+b^2)^2}{2}$
$(a^2+b^2)(1+1)\geq (a+b)^2=100$

$\Rightarrow a^2+b^2\geq 50$

$\Rightarrow C\geq \frac{50^2}{2}=1250$

Vậy $C_{\min}=1250$

Giá trị này đạt tại $a=b=5\Leftrightarrow x=2$

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

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

Do \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left(x-2\right)^2+1>0\end{matrix}\right.\) ;\(\forall x\)

\(\Rightarrow A\ge0\) ;\(\forall x\)

\(A_{min}=0\) khi \(x=-1\)

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=x^2-4x+y^2-y+5`

`=(x^2-4x+4)+(y^2-y+1/4)+3/4`

`=(x-2)^2+(y-1/2)^2+3/4`

`=>A_(min) = 3/4 <=> {(x=2),(y=-1/2):}`

Ta có: \(\sqrt{x^2-2x+10}=\sqrt{x^2-2x+1+9}=\sqrt{\left(x-1\right)^2+9}\ge\sqrt{9}\ge3\)

          \(\sqrt{x^2+4x+5}=\sqrt{x^2+4x+4+1}=\sqrt{\left(x+2\right)^2+1}\ge\sqrt{1}\ge1\)

    \(\Rightarrow\)   \(\sqrt{x^2-2x+10}+\sqrt{x^2+4x+5}\ge1+3\ge4\)

Vậy GTNN của biểu thức là 4

a) \(A=x^2-2x+5\)

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

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

Vì \(\left(x-1\right)^2\ge0;\forall x\)

\(\Rightarrow\left(x-1\right)^2+4\ge0;\forall x\)

b) a sẽ làm tắt 1 vài bước nhé khi nào kiểm tra thì em làm theo mẫu a là được 

\(B=4x^2+4x+11\)

\(=4\left(x^2+x+\frac{11}{4}\right)\)

\(=4\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+\frac{11}{4}\right)\)

\(=4\left[\left(x+\frac{1}{2}\right)^2+\frac{10}{4}\right]\)

\(=4\left(x+\frac{1}{2}\right)^2+10\ge10;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(B_{min}=10\Leftrightarrow x=\frac{-1}{2}\)

c) Tìm GTLN nhé 

 \(C=5-8x-x^2\)

\(=-x^2-2.x.4-16+16+5\)

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

Vì \(-\left(x+4\right)^2\le0;\forall x\)

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

Dấu "="xảy ra\(\Leftrightarrow\left(x+4\right)^2=0\)

                     \(\Leftrightarrow x=-4\)

Vậy\(C_{max}=21\Leftrightarrow x=-4\)

A = x² - 2x + 5

= ( x² - 2x + 1 ) + 4

= ( x - 1 )² + 4 ≥ 4 > 0 ∀ x ( đpcm )

B = 4x² + 4x + 11

= ( 4x² + 4x + 1 ) + 10

= ( 2x + 1 )² + 10 ≥ 10 ∀ x

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

=> MinB = 10 <=> x = -1/2

C = 5 - 8x - x²

= -( x² + 8x + 16 ) + 21

= -( x + 4 )² + 21 ≤ 21 ∀ x

Đẳng thức xảy ra <=> x + 4 = 0 => x = -4

=> MaxC = 21 <=> x = -4

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

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

\(=\left(x-1\right)^2+4\ge4>0\forall x\)

\(\Rightarrowđpcm\)

\(B=4x^2+4x+11\)

\(=\left[\left(2x\right)^2+2.2x+1\right]+10\)

\(=\left(2x+1\right)^2+10\ge10\forall x\)

Dấu"="xảy ra khi \(\left(2x+1\right)^2=0\Rightarrow x=\frac{-1}{2}\)

\(Min_B=10\Leftrightarrow x=\frac{-1}{2}\)

Bài 5:

a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)

\(minA=5\Leftrightarrow x=2\)

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

\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Bài 4:

a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxM=7\Leftrightarrow x=2\)

b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)

a) x² - 2x + 5 = (x - 1)² + 4 >= 4

Min là 4 khi x = 1