Dạ, ĐK: \(n,a\inℕ^∗\)bn nhé !
\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}\)
\(=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
Do đó : \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)
Mk cảm ơn bn nhé :))