Question

Chứng minh rằng 

\(\dfrac{k}{n.\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\left(n;kEN^{\cdot}\right)\)

Answer 1

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

Answer 2

\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{n+k-n}{n\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)(đpcm)