\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\\ =\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\\ =\left(x^2+5x+4\right)\left(x^2+5x+6\right)+\)
Đặt \(x^2+5x+5=t\)
\(\Rightarrow A=\left(t-1\right)\left(t+1\right)+1=t^2\ge0\)
Dấu "=" xảy ra khi \(x^2+5x+5=0\)
Min A =0 khi x^2+5x+5=0