Lời giải:

$|x-2|\geq 0$ với mọi $x\in\mathbb{R}$

$|y+1|\geq 0$ với mọi $y\in\mathbb{R}$

$\Rightarrow A\geq 0+0-5=-5$

Vậy $A_{\min}=-5$. Giá trị này đạt tại $x-2=y+1=0$

$\Leftrightarrow x=2; y=-1$

$A$ không có max bạn nhé.

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

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 \(x=-\dfrac{1}{2}\)

a: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2< =2\forall x\)

Dấu '=' xảy ra khi x+2=0

=>x=-2

b: \(B=-2x^2-3x+5\)

\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}< =\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{3}{4}=0\)

=>\(x=-\dfrac{3}{4}\)

c: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

\(=-x^2-2x-1+9\)

\(=-\left(x^2+2x+1\right)+9\)

\(=-\left(x+1\right)^2+9< =9\forall x\)

Dấu '=' xảy ra khi x+1=0

=>x=-1

d: \(D=-8x^2+4xy-y^2+3\)

\(=-8\left(x^2-\dfrac{1}{2}xy\right)-y^2+3\)

\(=-8\left(x^2-2\cdot x\cdot\dfrac{1}{4}y+\dfrac{1}{16}y^2\right)+\dfrac{1}{2}y^2-y^2+3\)

\(=-8\left(x-\dfrac{1}{4}y\right)^2-y^2+3< =3\forall x,y\)

Dấu '=' xảy ra khi y=0 và x-1/4y=0

=>y=0 và x=0

c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)

Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)

Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)