\(a,\pi\approx3\)

\(b,\pi\approx3,14\)

\(c,\pi\approx3,1415\)

\(a,n^2+n+1⋮n+1\)

Ta có : \(n^2+n+1=n\left(n+1\right)+1\)

Vì \(n\left(n+1\right)⋮n+1\) nên để \(n^2+n+1⋮n+1\) thì \(1⋮n+1\)

\(\Rightarrow n\in\left\{1\right\}\)

\(b,7n⋮n-3\) mk chưa có thời gian làm

\(n^2+n+1⋮n+1\)

\(\left(25^9.11-25^2.25^7\right):28\)

\(=\left(25^9.11-25^9\right):25^8\)

\(=\left[25^9\left(11-1\right)\right]:25^8\)

\(=\left(25^9.10\right):25^8\)

\(=25^9:25^8.10\)

\(=25.10\)

\(=250\)