a: \(A=x^2-2x+1+y^2+4y+4+3\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+3>=3\)

Dấu '=' xảy ra khi x=1 và y=-2

b: \(B=x^2-4x+4+y^2-8y+16-14\)

\(=\left(x-2\right)^2+\left(y-4\right)^2-14>=-14\)

Dấu '=' xảy ra khi x=2 và y=4

a) \(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(A=\left(x^2+5x\right)^2-6^2\)

\(A=\left(x^2+5x\right)^2-36\)

Vì \(\left(x^2+5x\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)

\(\Rightarrow Amin=-36\Leftrightarrow x^2+5x=0\)

\(\Rightarrow x\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

b) \(B=x^2-2x+y^2+4y+8\)

\(B=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\)

\(B=\left(x-1\right)^2+\left(y+2\right)^2+3\)

Vì \(\left(x-1\right)^2\ge0\) với mọi x

\(\left(y+2\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\) với mọi x,y

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\)

\(\Rightarrow Bmin=3\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

c) \(C=x^2-4x+y^2-8y+6\)

\(C=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)

\(C=\left(x-2\right)^2+\left(y-4\right)^2-14\)

Vì \(\left(x-2\right)^2\ge0\) với mọi x

\(\left(y-4\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\) với mọi x,y

\(\Rightarrow Cmin=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

a) \(A=x^2+6x+11\)

\(A=x^2+6x+9+2\)

\(A=\left(x+3\right)^2+2\)

Có: \(\left(x+3\right)^2\ge0\Rightarrow\left(x+3\right)^2+2\ge2\)

Dấu = xảy ra khi: \(\left(x+3\right)^2=0\Rightarrow x+3=0\Rightarrow x=-3\)

Vậy: \(Min_A=2\) tại \(x=-3\)

b) \(B=4x-x^2+1\)

\(B=-x^2+4x-4+5\)

\(B=-\left(x-2\right)^2+5\)

\(B=5-\left(x-2\right)^2\)

Có: \(\left(x-2\right)^2\ge0\)

\(\Rightarrow5-\left(x-2\right)^2\le5\)

Dấu = xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

Vậy: \(Max_B=5\) tại \(x=2\)

d, (x-1) (x+2) (x+3) (x+6)
=(x^2+2x-x-2) (x^2+6x+3x+18)
=(x^2-x^2) + (2x-x+6x-3x) = (-2+18)
=0            + (-8x)              =16
=                    x                =16:(-8)
=                  x                  =-2

A = x^2 + 6x + 11

= x^2 + 6x + 9 + 2

= (x + 3)^2 + 2

min = 2

a, A = x^2 + 6x + 11

= x^2 + 6x + 9 + 2

= (x + 3)^2 + 2

làm tiếp

b, x^2 - 20x + 101

= x^2  20x + 100 + 1

= (x - 10)^2 + 1

có (x - 10)^2 > 0 => (x - 10)^2 +  > 1

a/ Ta có:

\(A=x^2-6x+11\)

\(A=x\cdot x-3x-3x+3\cdot3+2\)

\(A=x\left(x-3\right)-3\left(x-3\right)+2\)

\(A=\left(x-3\right)\left(x-3\right)+2\)

\(A=\left(x-3\right)^2+2\)

Vì \(\left(x-3\right)^2\ge0\)

Nên GTNN của \(\left(x-3\right)^2\)là 0

=> \(A_{min}=0+2=2\)

mình chỉ biết a. thôi

a) ta có : \(A=x^2-6x+11\)

\(A=x.x-3x-3x+3.3+2\)

\(A=x\left(x-3\right)-3\left(x-3\right)+2\)

\(A=\left(x-3\right)\left(x-3\right)+2\)

\(A=\left(x-3\right)^2+2\)

vì \(\left(x-3\right)^2\ge0\)

nên GTNN của \(\left(x-3\right)^2\)là \(0\)

\(\Rightarrow\)\(A_{min}\)\(=0+2=2\)

oOo Không đủ can đảm để oOo copy mà nói nhưu mk tự làm

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

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

\(=\left(x-2y-3\right)\left(x+2y\right)\)

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

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

c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)

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

d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)

Bài 1:

\(A=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1\)

\(A_{min}=1\) khi \(x+1=0\Leftrightarrow x=-1\)

\(B=\left(x-3\right)^2\ge0\)

\(B_{min}=0\) khi \(x=3\)

\(C=2\left(x^2-2.\frac{3}{2}x+\frac{9}{4}\right)+\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2+\frac{9}{2}\ge\frac{9}{2}\)

\(C_{min}=\frac{9}{2}\) khi \(x=\frac{3}{2}\)

\(D=\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)

\(D=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(D_{min}=\frac{3}{4}\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=-3\end{matrix}\right.\)

Bài 2:

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

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

\(B=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(B_{max}=7\) khi \(x=2\)

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

\(C_{max}=\frac{1}{4}\) khi \(x=\frac{1}{2}\)

\(D=-\left(x^2-2x+1\right)-\left(y^2-4y+4\right)+11\)

\(D=-\left(x-1\right)^2-\left(y-2\right)^2+11\le11\)

\(D_{max}=11\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(E=-\frac{1}{2}\left(4x^2-4x+1\right)-\frac{9}{2}=-\frac{1}{2}\left(2x-1\right)^2-\frac{9}{2}\le-\frac{9}{2}\)

\(E_{max}=-\frac{9}{2}\) khi \(x=\frac{1}{2}\)