1)

\(A=n^5-n=n\left(n^4-1\right)=n\left(n^2-1\right)\left(n^2+1\right)\)

Nếu n chia hết hết cho 2 thì A chia hết cho 2

Nếu n không chia hết cho 2 thì \(n^2-1\) chia hết cho 2 => A chia hết cho 2

CMTT : A chia hết cho 5. Gợi ý : SCP chia 5 dư 0;1;4

2)

\(x^7+x^2+1\\ =x^7-x+x^2+x+1\\ =x\left(x^6-1\right)+\left(x^2+x+1\right)\\ =x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\\ =x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x\left(x-1\right)\left(x^3+1\right)+1\right)\\ =\left(x^2+x+1\right)\left(\left(x^2-x\right)\left(x^3+1\right)+1\right)\\ =\left(x^2+x+1\right)\left(x^5+x^2-x^4-x+1\right)\\ =\left(x^2+x+1\right)\left(\left(x^5-x^4\right)+\left(x^2-x\right)+1\right)\\ =\left(x^2+x+1\right)\left(x^4\left(x-1\right)+x\left(x-1\right)+1\right)\\ =\left(x^2+x+1\right)\left(\left(x-1\right)\left(x^4+x\right)+1\right)\)