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

Dấu = xảy ra khi x=-7

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

Dấu = xảy ra khi x=-2

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

Dấu = xảy ra khi x=1/2

D=(x+3/2)^2+3/4>=3/4

Dấu = xảy ra khi x=-3/2

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

Dấu = xảy ra khi x=-1

A = x² + 14x + 50 = (x² + 14x + 49) + 1 = (x + 7)² + 1

Ta có: (x + 7)² \(\ge\)0 \(\forall\)x

=> (x + 7)² + 1 \(\ge\)1 \(\forall\)x

Dấu "=" xảy ra <=> x + 7 = 0 <=> x = -7

Vậy MinA = 1 <=> x = -7

B = x² + 4x + 6 = (x² + 4x + 4) + 2 = (x + 2)² + 2

Ta luôn có : (x + 2)² \(\ge\)0 \(\forall\)x

=> (x + 2)² + 2 \(\ge\)2 \(\forall\)x

Dấu "=" xảy ra <=> x + 2 = 0 <=> x = -2

Vậy MinB = 2 <=> x = -2

C = x² - x + 1 = (x² - x + 1/4) + 3/4 = (x - 1/2)² + 3/4

Ta luôn có: (x - 1/2)² \(\ge\)0 \(\forall\)x

=> (x - 1/2)² + 3/4 \(\ge\)3/4 \(\forall\)x

Dấu "=" xảy ra <=> x - 1/2 = 0 <=> x = 1/2

Vậy MinC = 3/4 <=> x = 1/2

A = x2 + 14x + 50 = (x2 + 14x + 49) + 1 = (x + 7)2 + 1

Ta có: (x + 7)2 \ge≥0 \forall∀x

=> (x + 7)2 + 1 \ge≥1 \forall∀x

Dấu "=" xảy ra <=> x + 7 = 0 <=> x = -7

Vậy Min A = 1 <=> x = -7

B = x2 + 4x + 6 = (x2 + 4x + 4) + 2 = (x + 2)2 + 2

Ta luôn có : (x + 2)2 \ge≥0 \forall∀x

=> (x + 2)2 + 2 \ge≥2 \forall∀x

Dấu "=" xảy ra <=> x + 2 = 0 <=> x = -2

Vậy Min B = 2 <=> x = -2

C = x2 - x + 1 = (x2 - x + 1/4) + 3/4 = (x - 1/2)2 + 3/4

Ta luôn có: (x - 1/2)2 \ge≥0 \forall∀x

=> (x - 1/2)2 + 3/4 \ge≥3/4 \forall∀x

Dấu "=" xảy ra <=> x - 1/2 = 0 <=> x = 1/2

Vậy Min C = 3/4 <=> x = 1/2

D:E ko biết làm chúc bạn học tốt

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 $a=b=5\Leftrightarrow x=2$

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.

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

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 \(x=-1\)

\(a,A=\left(x^2-x\right)\left(x^2-x-12\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)+36-36\\ A=\left(x^2-x+6\right)^2-36\ge-36\\ A_{min}=-36\Leftrightarrow x^2-x+6=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,B=4x^4+4x^3+5x^2+4x+3\\ B=\left(4x^4+4x^3+x^2\right)+\left(x^2+4x+4\right)-1\\ B=x^2\left(2x+1\right)^2+\left(x+2\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow\left\{{}\begin{matrix}x\left(2x+1\right)=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy dấu \("="\) không xảy ra