Lời giải:
Ta có: \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}\)
\(=\dfrac{\left(n+a\right)-n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)
Vậy, \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)