Bài 13:

1: \(A=-x^2+4x+3\)

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

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

Dấu '=' xảy ra khi x=2

2: \(B=-\left(x^2-6x+11\right)\)

\(=-\left(x-3\right)^2-2\le-2\)

Dấu '=' xảy ra khi x=3

mình biết mỗi bài 4:

A={2007}

mình đi xin bn đó

cảm ơn bạn Xử Nữ các bạn khác giúp mình với

1. n=6
2. k thể làm đc
3. n=3
4. n=2
5. ko bik làm xin lỗi nhiều!
6. n=2
7. n=4
8. n=1

Bài 2:

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

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

\(=x^4-2x^3+x^2+2x^2-4x-1\)

\(=x^4-2x^3+3x^2-4x-1\)

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

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

Dấu '=' xảy ra khi \(\begin{cases}x^2-x=0\\ x-1=0\end{cases}\Rightarrow x=1\)

b: \(B=\left(x-5\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+2022\)

\(=\left(x-5\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)+2022\)

\(=\left(x^2-x-20\right)\left(x^2-x-6\right)+2022\)

\(=\left(x^2-x-6\right)^2-14\left(x^2-x-6\right)+49+1973=\left(x^2-x-6+7\right)^2+1973\)

\(=\left(x^2-x+1\right)^2+1973\)

Ta có: \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34\forall x\)

=>\(\left(x^2-x+1\right)^2\ge\frac{9}{16}\forall x\)

=>\(\left(x^2-x+1\right)^2+1973\ge\frac{9}{16}+1973\forall x\)

=>B>=31577/16∀x

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

\(x^3-9x^2+26x-24\)

\(=x^3-4x^2-5x^2+20x+6x-24\)

\(=\left(x-4\right)\left(x^2-5x+6\right)\)

\(=\left(x-4\right)\left(x-2\right)\left(x-3\right)\)

Bài 2:

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

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

\(=x^4-2x^3+x^2+2x^2-4x-1\)

\(=x^4-2x^3+3x^2-4x-1\)

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

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

Dấu '=' xảy ra khi \(\begin{cases}x^2-x=0\\ x-1=0\end{cases}\Rightarrow x=1\)

b: \(B=\left(x-5\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+2022\)

\(=\left(x-5\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)+2022\)

\(=\left(x^2-x-20\right)\left(x^2-x-6\right)+2022\)

\(=\left(x^2-x-6\right)^2-14\left(x^2-x-6\right)+49+1973=\left(x^2-x-6+7\right)^2+1973\)

\(=\left(x^2-x+1\right)^2+1973\)

Ta có: \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34\forall x\)

=>\(\left(x^2-x+1\right)^2\ge\frac{9}{16}\forall x\)

=>\(\left(x^2-x+1\right)^2+1973\ge\frac{9}{16}+1973\forall x\)

=>B>=31577/16∀x

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

Bài 2:

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

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

\(=x^4-2x^3+x^2+2x^2-4x-1\)

\(=x^4-2x^3+3x^2-4x-1\)

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

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

Dấu '=' xảy ra khi \(\begin{cases}x^2-x=0\\ x-1=0\end{cases}\Rightarrow x=1\)

b: \(B=\left(x-5\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+2022\)

\(=\left(x-5\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)+2022\)

\(=\left(x^2-x-20\right)\left(x^2-x-6\right)+2022\)

\(=\left(x^2-x-6\right)^2-14\left(x^2-x-6\right)+49+1973=\left(x^2-x-6+7\right)^2+1973\)

\(=\left(x^2-x+1\right)^2+1973\)

Ta có: \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34\forall x\)

=>\(\left(x^2-x+1\right)^2\ge\frac{9}{16}\forall x\)

=>\(\left(x^2-x+1\right)^2+1973\ge\frac{9}{16}+1973\forall x\)

=>B>=31577/16∀x

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)