Question

CMR:  \(\dfrac{a}{n\left(n+a\right)}\)=\(\dfrac{1}{n}\) - \(\dfrac{1}{n+a}\) (n,a ϵ N^*)

Answer 1

\(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)

\(\dfrac{a}{n+\left(n+a\right)}+\dfrac{1}{n+a}=\dfrac{1}{n}\)

Vậy ta sẽ CRM\(\dfrac{a}{n+\left(n+a\right)}+\dfrac{1}{n+a}=\dfrac{1}{n}\)

\(\dfrac{a}{n\left(n+a\right)}+\dfrac{1}{n+a}\)

\(=\dfrac{a}{n}\cdot\dfrac{1}{\left(n+a\right)}+\dfrac{1}{n+a}\)

\(=\dfrac{1}{n+a}\cdot\left(\dfrac{a}{n}+1\right)\)

\(=\dfrac{1}{n+a}\cdot\dfrac{a+n}{n}\)

Đã \(CMR:\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)