\(x^3-19x-30=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)

\(x^3-19x-30\)

\(=x^3+2x^2-2x^2-4x-15x-30\)

\(=x^2\left(x+2\right)-2x\left(x+2\right)-15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-2x-15\right)\)

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

\(x^3-3x^2-6x+8\\ =\left(x^3+8\right)-3x\left(x+2\right)\\ =\left(x+2\right)\left(x^2-2x+4\right)-3x\left(x+2\right)\\ =\left(x+2\right)\left(x^2-5x+4\right)\\ =\left(x+2\right)\left(x-4\right)\left(x-1\right)\)

bước 3 đến 4 là sao mình ko hiểu nhỉ

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

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)

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

3x² + 9x - 30

<=> 3x² + 15x - 6x -30

<=> 3x (x+5) - 6 (x+5)

<=> (3x - 6) + (x+5)

\(a,2x+2y-x^2-xy=2x+2y-\left(x^2+xy\right).\)

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

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

\(a,2x+2y-x^2-xy=2.\left(x+y\right)-x.\left(x+y\right)=\left(2-x\right).\left(x+y\right)\)

\(b,4x^3+9x^2-19x-30=3x^3+x^3+9x^2-9x-10x-30\)

\(=3x^2.\left(x+3\right)+x.\left(x-3\right).\left(x+3\right)-10.\left(x+3\right)=\left(x+3\right).\left(3x^2+x^2-3x-10\right)\)

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 \(x-\frac12=0\)

=>\(x=\frac12\)