Question

cho 2 phân số \(\frac{1}{n}\) và \(\frac{1}{n+1}\)(n∈Z) . Chứng tỏ rằng

\(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n}-\frac{1}{n-1}\)

Answer 1

Ta có

\(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n\left(n+1\right)};\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)

vậy \(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n}-\frac{1}{n+1}\)

Answer 2

Ta có: \(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

Vậy \(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n}-\frac{1}{n-1}\) (ĐPCM)

Answer 3

Ta có:

Vậy (ĐPCM)