Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

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

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

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

(x²+x+1)(x²+x+2)-12

=(x²+x+1)[(x²+x+1)+1)-12

=(x²+x+1)²+(x²+x+1)-12

=(x²+x+1)²-3.(x²+x+1)+4.(x²+x+1)-12

=(x²+x+1)(x²+x+1-3)+4.(x²+x+1-3)

=(x²+x+1)(x²+x-2)+4.(x²+x-2)

=(x²+x-2)(x²+x+1+4)

=(x²-x+2x-2)(x²+x+5)

=[x.(x-1)+2.(x-1)](x²+x+5)

=(x-1)(x+2)(x²+x+5)

(x^2+x+1)(x^2+x+2)-12

Đặt x^2+x+1= a ta có

=a^2+a-12

=a^2-3a+4a-12

=(a^2-3a)+(4a-12)

=a(a-3)+4(a-3)

=(a-3)(a+4)

thay x^2+x+1=a ta được

(x^2+x-2)(x^2+x+5)

                                         Bài làm :

Ta có :

(x²+x+1)(x²+x+2)-12

=(x²+x+1)[(x²+x+1)+1)-12

=(x²+x+1)²+(x²+x+1)-12

=(x²+x+1)²-3.(x²+x+1)+4.(x²+x+1)-12

=(x²+x+1)(x²+x+1-3)+4.(x²+x+1-3)

=(x²+x+1)(x²+x-2)+4.(x²+x-2)

=(x²+x-2)(x²+x+1+4)

=(x²-x+2x-2)(x²+x+5)

=[x.(x-1)+2.(x-1)](x²+x+5)

=(x-1)(x+2)(x²+x+5)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

Đặt x^2 + x +1 = a 

Thay vào ta có :

a(a+1) - 12 

= a^2 + a - 12 

= a^2 + 4a - 3a - 12 

= a(a+4 ) - 3 (a + 4 )

=(a- 3 )(a+4 )

Thay a = x^2 + x + 1 ta có :

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

=(x^2 + x - 2 )(x^2 + x + 5 )

\(x^2-x-12\\ =x^2-4x+3x-12\\ =x\left(x-4\right)+3\left(x-4\right)\\ =\left(x-4\right)\left(x+3\right)\)

(x^2+3x+2)(x^2+7x+12)+1

=(x²+x+2x+2)(x²+3x+4x+12)+1

=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]+1

=(x+1)(x+2)(x+3)(x+4)+1

=[(x+1)(x+4)][(x+2)(x+3)]+1

=(x²+5x+4)(x²+5x+6)+1

=(x²+5x+4)[(x²+5x+4)+2]+1

=(x²+5x+4)²+2(x²+5x+4)+1

=(x²+5x+4+1)²

=(x²+5x+5)²

\(=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16\\ =\left(x^2+x+2\right)^2-16\\ =\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\\ =\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

=\(x^4+2x^3+x^2+4x^2+4x-12\)

=\(x^4+2x^3+5x^2+4x-12\)

=\(x^4-x^3+3x^3-3x^2+8x^2+4x-12\)

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

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

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

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

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

=\((x-1)(x^3+3x^2+8x+12)\)