a: Sửa đề: \(\dfrac{\left(3x^3+10x^2-5\right)}{3x+1}\)
\(=\dfrac{3x^3+x^2+9x^2+3x-3x-1-4}{3x+1}\)
\(=x^2+3x-1+\dfrac{-4}{3x+1}\)
b: \(3n^3+10n^2-5⋮3n+1\)
=>\(3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)
=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;-1;1\right\}\)