\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(=\left(x^2+3x\right)\left(x^2+2x+x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt : \(x^2+3x+1=y\)
\(\Leftrightarrow\) Đa thức có dạng :
\(\left(y-1\right)\left(y+1\right)+1\)
\(=y^2+y-y-1-1\)
\(=y^2-2\)
\(=\left(y-\sqrt{2}\right)\left(y+\sqrt{2}\right)\)
Thay \(y=x^2+3x+1\) ta được :
\(\left(x^2+3x+1-\sqrt{2}\right)\left(x^2+3x-1+\sqrt{2}\right)\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
