\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)(1)
Đặt \(x^2+3x+1=t\)thay vào (1) ta được :
\(\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1\)
\(=t^2\)Thay \(t=x^2+3x+1\)ta được:
\(\left(x^2+3x+1\right)^2\)
\(=\left(x^2+2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+1\right)^2\)
\(=\left[\left(x+\frac{3}{2}\right)-\frac{5}{4}\right]^2\)
\(=\left(x+\frac{3}{2}-\frac{\sqrt{5}}{2}\right)^2\left(x+\frac{3}{2}+\frac{\sqrt{5}}{2}\right)^2\)
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