2.a) (ko phân tích được, bạn coi lại nhé)

b) phần còn lại của chứng minh là gì thế bạn?

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 \(x-\frac12=0\)

=>\(x=\frac12\)

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 \(x-\frac12=0\)

=>\(x=\frac12\)

\(=x^3-x+7x+7=x\left(x-1\right)\left(x+1\right)+7\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x+7\right)\)

Sửa đề: x^3+6x^2+11x+6

=x^3+x^2+5x^2+5x+6x+6

=(x+1)(x^2+5x+6)

=(x+1)(x+2)(x+3)

x²-4+(x-2)²=(x+2)(x-2)+(x-2)(x-2)=(x-2)(x+2+x-2)=(x-2).2x

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

\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)

\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)

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

\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)

b: =xy-x-y+1

=x(y-1)-(y-1)

=(x-1)(y-1)

c: =(x-2y)^2-4y

\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)

d: =16-(x^2-2xy+y^2)

=16-(x-y)^2

=(4-x+y)(4+x-y)