Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

het thoirui pan oi

a: Ta có: \(x^5-x^3+x^2-1\)

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

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

b: Ta có: \(5x^3-45x\)

\(=5x\left(x^2-9\right)\)

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

c: Ta có: \(16x^4y^2+2xy^5\)

\(=2xy^2\left(8x^3+y^3\right)\)

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

d: Ta có: \(a^3-8+6a^2-12a\)

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

\(=\left(a-2\right)\left(a^2+8a+4\right)\)

e: Ta có: \(x^4+x^3+x+1\)

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

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

a)  \(A=\left(x+2\right)\left(x+3\right)\left(x+5\right)\left(x+6\right)-10\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+15\right)-10\)

Đặt   \(x^2+8x+12=t\)

Khi đó ta có: 

\(A=t\left(t+3\right)-10\)

   \(=t^2+3t-10\)

   \(=\left(t-2\right)\left(t+5\right)\)

Thay trở lại ta có:

\(A=\left(x^2+8x+10\right)\left(x^2+8x+17\right)\)

b)  \(B=x\left(2x+1\right)\left(2x+3\right)\left(4x+8\right)-18\)

\(=\left(4x^2+8x\right)\left(4x^2+8x+3\right)-18\)

Đặt  \(4x^2+8x=t\)

Khi đó ta có:

\(B=t\left(t+3\right)-18=t^2+3t-18=\left(t-3\right)\left(t+6\right)\)

Thay trở lại ta có:

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

Mọi người đã hướng dẫn bạn cách làm rồi mà.

a, Đặt A=...=(x+2)(x+6)(x+3)(x+5)-10=(x²+8x+12)(x²+8x+15)-10

Đặt x²+8x+12=y

=>A=y(y+3)-10=y²+3y-10=y²-2y+5y-10=y(y-2)+5(y-2)=(y-2)(y+5)=(x²+8x+12-2)(x²+8x+12+5)=(x²+8x+10)(x²+8x+17)

b, Đặt B=...=x(4x+8)(2x+1)(2x+3)-18=(4x²+8x)(4x²+8x+3)-18

Đặt 4x²+8x=t

=>B=t(t+3)-18=t²+3t-18=t²-3t+6t-18=t(t-3)+6(t-3)=(t-3)(t+6)=(4x²+8x-3)(4x²+8x+6)

Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

b: =>(x+3)(x-3)=0

=>x=-3 hoặc x=3

 a) 3x2 – 7x + 2

\(=3x^2-6x-x+2\)

\(=\left(3x^2-6x\right)-\left(x-2\right)\)

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

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

 b) a(x2 + 1) – x(a2 + 1)

\(=ax^2+a-\left(a^2x+x\right)\)

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

.......?

a) Ta có: \(3x^2-7x+2\)

\(=3x^2-6x-x+2\)

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

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

b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)

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

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

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)