\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
= \(\left(x^2+3x\right)\left(x^2+3x+2\right)+1\) (*)
Đặt \(x^2+3x=a\)
Pt (*) \(\Leftrightarrow a.\left(a+2\right)+1\) = \(a^2+2a+1\) = \(\left(a+1\right)^2\)
= \(\left(x^2+3x+1\right)^2\)
Sửa đề:
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+2\)
\(=\left[x.\left(x+3\right)\right].\left[\left(x+1\right)\left(x+2\right)\right]+2\)
\(=\left(x^2+3x\right).\left(x^2+2x+x+3\right)+2\)
\(=\left(x^2+3x\right)\left(x^2+3x+3\right)+2\)(1)
Đặt \(x^2+3x=a\Rightarrow x^2+3x+3=a+3\)
Từ đó suy ra:
\(\left(1\right)=a.\left(a+3\right)+2=a^2+3a+2=a^2+a+2a+2\)
\(=a.\left(a+1\right)+2.\left(a+1\right)=\left(a+1\right).\left(a+2\right)\)(*)
Vì \(a=x^2+3x\) nên
(*)\(=\left(x^2+3x+1\right)\left(x^2+3x+2\right)\)
\(=\left(x^2+3x+1\right)\left(x^2+x+2x+2\right)\)
\(=\left(x^2+3x+1\right)\left[x.\left(x+1\right)+2.\left(x+1\right)\right]\)
\(=\left(x^2+3x+1\right)\left(x+1\right)\left(x+2\right)\)
Chúc bạn học tốt!!!