ta có
\(A=10n^2+n-10=10n^2-10n+11n-11+1=\left(10n^2-10n\right)+\left(11n-11\right)+1\)
\(=10n\left(n-1\right)+11\left(n-1\right)+1=\left(n-1\right)\left(10n+11\right)+1\)
DO \(\left(n-1\right)\left(10n+11\right)⋮\left(n-1\right)\)nên để A chia hết cho n-1 thì \(1⋮\left(n-1\right)\Rightarrow n-1\inƯ\left(1\right)=\left\{1;-1\right\}\Rightarrow n\in\left\{2;0\right\}\)