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

ĐK: \(-3\le x\le2\)

\(4\left(x+1\right)\left(\sqrt{x+3}-\sqrt{2-x}\right)=-x^2+12x+13\)

<=> \(4\left(x+1\right)\left(\sqrt{x+3}-\sqrt{2-x}\right)+\left(x+1\right)\left(x-13\right)=0\)

<=> \(\left(x+1\right)\left[4\left(\sqrt{x+3}-\sqrt{2-x}\right)+x-13\right]=0\)

<=> \(\orbr{\begin{cases}x+1=0\left(1\right)\\4\left(\sqrt{x+3}-\sqrt{2-x}\right)+x-13=0\left(2\right)\end{cases}}\)

(1) <=> x = - 1 ( thỏa mãn ) 

(2) <=> \(4\left(\sqrt{x+3}-\sqrt{2-x}\right)=13-x\)

Ta có VT \(\le4\sqrt{x+3+2-x}=4\sqrt{5}\)với \(-3\le x\le2\)

\(VP\ge11\)với \(-3\le x\le2\)

=> VP > VT mọi \(-3\le x\le2\)

pt (2) vô nghiệm 

Vậy x = - 1 là nghiệm.