Cho hỏi cái "Cho abc=1" để làm gì thế:v?
Ta có: \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)\(=\left(x^2+3x+1-1\right)\left(x^2+3x+1+1\right)+1\)
\(=\left(x^2+3x+1\right)^2-1^2+1=\left(x^2+3x+1\right)^2\)
Ta thấy: \(\left(x^2+3x+1\right)^2\ge0\) (Với mọi x)
\(\Rightarrow x\left(x+1\right)\left(x+2\right)\left(x+3\right)\ge0\)
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