a) đặt \(x^2+x=y\)
\(\left(x^2+x\right)^2+3\left(x^2+x\right)+2=y^2+3y+2=\left(y+1\right)\left(y+2\right)=\left(x^2+x+1\right)\left(x^2+x+2\right)\)
b) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(x^2+3x+1=y\)
\(\left(x^2+3x\right)\left(x^2+3x+2\right)+1=\left(y-1\right)\left(y+1\right)+1=y^2=\left(x^2+3x\right)^2=x^2\left(x+3\right)^2\)
c) \(\left(x^2+x+1\right)\left(x^2+3x+1\right)+x^2=x^4+4x^3+6x^2+4x+1=\left(x^4+2x^3+x^2\right)+\left(2x^3+4x^2+2x\right)+\left(x^2+2x+1\right)=\left(x+1\right)^4\)
