Ta có:
\(3^{n+2}+3^n-2^{n+2}-2^n\)
= \(3^n.3^2+3^n-2^n.2^2-2^n\)
= \(3^n.\left(3^2+1\right)-2^n.\left(2^2+1\right)\)
= \(3^n.10-2^n.5\)
= \(3^n.10-2^{n-1}.10\) = \(\left(3^n-2^{n-1}\right).10\) => chia hết cho 10
Nhớ cho mk nha bạn !
Ta có 3^n+2+3^n-2^n+2-2^n
=3^n+2+3^n-(2^n+2+2^n)
=3^n.(3^2+1)-2^n.(2^2+1)
=3^n.10-2^n.10
=3^n.10-2^(n-1).10=10.(3^n-2^(n-1)) chia hết cho 10(đpcm)