\(M=x^{2017}-x^{2013}=x^{2013}\left(x^4-1\right)=x^{2013}\left(x^2-1\right)\left(x^2+1\right)=x^{2012}.x\left(x-1\right)\left(x+1\right)\)
Do \(x\left(x-1\right)\left(x+1\right)\) là tích 3 số nguyên liên tiếp nên \(x\left(x-1\right)\left(x+1\right)⋮6\)(1)
Ta lại có : \(x\left(x-1\right)\left(x+1\right)\left(x^2+1\right)=x\left(x+1\right)\left(x-1\right)\left(x^2-4+5\right)\)
\(=\left(x-1\right)x\left(x+1\right)\left(x-2\right)\left(x+2\right)-5x\left(x+1\right)\left(x-1\right)\)
Vì \(\left(x-1\right)x\left(x+1\right)\left(x-2\right)\left(x+2\right)⋮5\)(tích 5 số nguyên LT)
Nên \(\left(x-1\right)x\left(x+1\right)\left(x-2\right)\left(x+2\right)-5x\left(x+1\right)\left(x-1\right)⋮5\)
=> M chia hết cho 5 (2)
Từ (1) ; (2) => M chia hết cho 30
