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

=(x²+x+2x+2)(x²+3x+4x+12)

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

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

(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

(x^2+3x+2)(x^2+7x+12)-24

=(x²+x+2x+2)(x²+3x+4x+12)-24

=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]-24

=(x+1)(x+2)(x+3)(x+4)-24

=(x+1)(x+4)(x+2)(x+3)-24

=(x²+5x+4)(x²+5x+6)-24

Đặt t=x²+5x+4 ta được:

t.(t+2)-24

=t²+2t-24

=t²-4t+6t-24

=t.(t-4)+6.(t-4)

=(t-4)(t+6)

thay t=x²+5x+4 ta được:

(x²+5x+4-4)(x²+5x+4+6)

=(x²+5x)(x²+5x+10)

=x.(x+5)(x²+5x+10)

Vậy (x^2+3x+2)(x^2+7x+12)-24=x.(x+5)(x²+5x+10)

hay quá bạn ơi

Đúng nhưng hơi dài

Đặt x^2+5x+5=t thì ta được

(t-1)(t+1)-24=t^2-1-24=t^2-25=(t-1)(t+1)

=(x^2+5x)(x^2+5x+10)=x(x+5)(x^2+5x+10)

Sẽ gọn hơn

Ta có :  \(M=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left[\left(x+1\right)\left(x+4\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]+1=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(t=x^2+5x+5\) \(\Rightarrow M=\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)

Vậy \(M=\left(x^2+5x+5\right)^2\)

\(\times^2+7\times+12\)

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

\(=\times\left(\times+4\right)+3\left(\times+4\right)\)

\(=\left(\times+4\right)\left(\times+3\right)\)

\(x^2+7x+12=x^2+3x+4x+12\)

                            \(=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)

\(x^2+7x+12\)

cách 1: \(=x^2+4x+3x+12\)

\(=x\left(x+4\right)+3\left(x+4\right)\)

\(=\left(x+4\right)\left(x+3\right)\)

cách 2: \(=x^2+3x+4x+12\)

\(=x\left(x+3\right)+4\left(x+3\right)\)

\(=\left(x+3\right)\left(x+4\right)\)

cách 3: \(=\left(x^2+7x+12,25\right)-0.25\)

\(=\left(x+3.5\right)^2-0.5^2\)

\(=\left(x+3.5+0.5\right)\left(x+3.5-0.5\right)\)

\(=\left(x+4\right)\left(x+3\right)\)

lấy đâu ra 8 cách vậy trời!!!!!!!!!!!!!!!

Cách 1: 

   \(x^2+7x+12\)

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

\(=x\left(x+4\right)+3\left(x+4\right)\)

\(=\left(x+3\right)\left(x+4\right)\)