(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

\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-20\)

\(=\left(x^2+5x+4\right)\cdot\left(x^2+5x+6\right)-20\)

Đặt:   \(x^2+5x+5=a\)Khi đó ta có:

\(A=\left(a-1\right)\left(a+1\right)-20=a^2-21=\left(a-\sqrt{21}\right)\left(a+\sqrt{21}\right)\)

tự thay trở lại

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 \(M=\left(x^2+5x+5\right)^2\)

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

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

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

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

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

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

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

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

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