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 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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

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

\(=x^4+2x^3+4x^2+3x-10\)

Đặt \(t=x^2+x+1\) thì ta có:

\(t\left(t+1\right)-12\)\(=t^2+t-12\)

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

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

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

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

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

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

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 )

đặt (x^2+x+1)=a

ta có : a(a+1)-12

     <=> a^2 +a -12

     <=> a^2 +4a - 3a -12

     <=> a(a+4) - 3(a+4)

     <=> (a+4)(a-3)

Đặt t=x²+x+1 ta được:

t.(t+1)-12

=t²+t-12

=t²-3t+4t-12

=t.(t-3)+4.(t-3)

=(t-3)(t+4)

thay t=x²+x+1 ta được:

(x²+x-2)(x²+x+5)

=(x²-x+2x-2)(x²+x+5)

=[x.(x-1)+2.(x-1)][x²+x+5)

=(x-1)(x+2)(x²+x+5)

Đặt x^2  +x + 1 = y thay vào ta có

 y(y+1) - 12 = y^2 + y - 12 = y^2  - 3y + 4y - 12 

 = y(y-3) + 4(y-3) = ( y- 3)( y + 4 ) 

Thay y   = 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+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