Ta có: \(a^5-a=a\left(a^2+1\right)\left(a^2-1\right)=a\left(a-1\right)\left(a+1\right)\left(a^2-4+5\right)\)

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

\(=a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)+5\left(a-1\right)\left(a+1\right)⋮5\)( 5 số nguyên liên tiếp chia hết cho 5)

=> \(a^5-a=a\left(a-1\right)\left(a+1\right)\left(a^2+1\right)⋮6\)

( 3 số nguyên liên tiếp chia hết cho 2 và chia hết cho 3 nên chia hết cho 6) 

mà 6 .5 = 30 ; ( 6;5) = 1 

=> \(a^5-a⋮30\)

=> \(a^{2020}-a^{2016}=a^{2015}\left(a^5-a\right)⋮30\)

=> \(A=\left(a^{2020}-a^{2016}\right)+\left(b^{2020}-b^{2016}\right)+\left(c^{2020}-c^{2016}\right)⋮30\)

A=21/B=32

Ta có: \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)

=> \(a+b\ge\frac{\left(a+b\right)^2}{2}\)

=> \(0< \left(a+b\right)\le2\)

=> \(P=\left(a^4+b^4\right)+\frac{2020}{\left(a+b\right)^2}\ge\frac{\left(a^2+b^2\right)^2}{2}+\frac{8}{\left(a+b\right)^2}+\frac{2012}{\left(a+b\right)^2}\)

\(=\frac{\left(a+b\right)^2}{2}+\frac{8}{\left(a+b\right)^2}+\frac{2012}{\left(a+b\right)^2}\)

\(\ge4+\frac{2012}{4}=507\)

Dấu "=" xảy ra <=> a = b = 1