a,Ta có:\(10^{2k}-1=10^{2k}-10^k+10^k-1=10^k.\left(10^k-1\right)+10^k-1=\left(10^k+1\right)\left(10^k-1\right)\) chia hết cho 9
b,Ta có:
\(10^{3k}-10^{2k}+10^{2k}-10^k+10^k-1=10^{2k}\left(10^k-1\right)+10^k\left(10^k-1\right)+10^k-1\)
\(=\left(10^{2k}+10^k+1\right).\left(10^k-1\right)\) chia hết cho 9