Question

biết( n,a thuộc n*)

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

Answer 1

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

Answer 2

Ta có :

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

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

Answer 3

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

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

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