Đặt \(x^2+x+1=t\) 

Ta có: \(\left(x^2+x+1\right)^2+3x\left(x^2+x+1\right)+2x^2\)

\(=t^2+3xt+2x^2\)

\(=t^2+xt+2xt+2x\)

\(=t\left(t+x\right)+2x\left(t+x\right)\)

\(=\left(t+x\right)\left(t+2x\right)\)

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

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

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

Chúc bạn học tốt.

Đặt x^2+2x=t =>3t^2-2t-1=3t^2-3t+t-1=3t(t-1)+(t-1)=(t-1)(3t+1)

=>(x^2+2x-1)(3x^2+6x+1)

\(Dat:a^2+a+1=b\Rightarrow....=a\left(a+1\right)-12=\left(a+4\right)\left(a-3\right)\) 

=

a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)   (1)

Đặt x² + x +1 = t 

Ta có : \(t\left(t+1\right)-12=t^2+t-12=t^2-3t+4t-12\)

\(=t\left(t-3\right)+4\left(t-3\right)=\left(t-3\right)\left(t+4\right)\)

Thay vào (1), ta được : \(\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)

b) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)  (2)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

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

Đặt x² + 7x + 11 = y

Ta có : \(\left(y-1\right)\left(y+1\right)-24=y^2-1-24=y^2-25=\left(y-5\right)\left(y+5\right)\)

Thay vào (2), ta được : \(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

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

\(=\left(x^2+6x\right)\left(x^2+6x+8\right)-9\)

\(=\left(x^2+6x\right)^2+8\left(x^2+6x\right)-9\)

\(=\left(x^2+6x+9\right)\left(x^2+6x-1\right)\)

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

Đặt x^2-3x-2=t =>(t+4)(t-4)+12=t-16+12=t-4=(t+2)(t-2)

=>(x^2-3x-2+2)(x^2-3x-2-2)=(x^2-3x)(x^2-3x-4)

\(Dat:x^2+x=a\Rightarrow....=a^2-2a-15=\left(a-1\right)^2-4^2=\left(a+3\right)\left(a-7\right)\) 

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

\(Dat:x+y=a\Rightarrow....=a^2-a-12=\left(a+3\right)\left(a-4\right)=\left(x+y+3\right)\left(x+y-4\right)\)

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

Đặt \(x^2+x=a\) .

Khi đó :  \(A=a^2-2a-15=a^2-5a+3a-15\)\(=a\left(a-5\right)+3\left(a-5\right)=\left(a+3\right)\left(a-5\right)\)

Mà \(a=x^2+x\) nên  \(A=\left(x^2+x+3\right)\left(x^2+x-5\right)\)

b) B = \(x^2+2xy+y^2-x-y-12\) \(=\left(x+y\right)^2-\left(x+y\right)-12\)

Đặt x+y = z.

Khi đó : \(B=z^2-z-12=z^2-4z+3z-12=z\left(z-4\right)+3\left(z-4\right)\)\(=\left(z+3\right)\left(z-4\right)\)

Mà z = x+y nên B = (x+y+3)(x+y-4)

Ta có : (x+2)(x+4)(x+6)(x+8) + 16

=[(x+2).(x+8)].[(x+4)(x+6)]+16

=(x²+10x+16).(x²+10x+24)+16 (1)

Đặt x^2+10x+16=a thì (1) trở thành:

a.(a+8)+16=a²+8a+16=(a+4)²=(x^2+10x+20)²