Giải:
a) \(n\left(2n-3\right)-2n\left(n+1\right):5\)
\(=2n^2-3n-2n^2+2n:5\)
\(=n:5=\dfrac{n}{5}\)
Vậy ...
b) \(n\left(n+3\right)-\left(n-3\right)\left(n+2\right):6\)
\(=n^2+3n-n^2-3n+2n-6:6\)
\(=2n-6:6\)
\(=2\left(n-3\right):6\)
\(=\left(n-3\right):3=\dfrac{n-3}{3}\)
Vậy ...