Question

chứng minh 

\(\frac{1}{n}\cdot\frac{1}{n+4}=\frac{1}{4}\cdot\left(\frac{1}{n}-\frac{1}{n+4}\right)\)

Answer 1

\(\frac{1}{n}.\frac{1}{n+4}=\frac{1}{n\left(n+4\right)}=\frac{1}{4}.\frac{4}{n\left(n+4\right)}=\frac{1}{4}.\frac{\left(n+4\right)-n}{n\left(n+4\right)}=\frac{1}{4}\left(\frac{1}{n}-\frac{1}{n+4}\right)\)

Vậy ta có đpcm

Answer 2

ta xét vế phải

A=\(\frac{1}{4}\).(\(\frac{1}{n}-\frac{1}{n+4}\))=\(\frac{1}{4}\).(\(\frac{n+4}{n.\left(n+4\right)}\)-\(\frac{n}{n.\left(n+4\right)}\))

=\(\frac{1}{4}\).\(\frac{4}{n.\left(n+4\right)}\)=\(\frac{1}{n.\left(n+4\right)}\)

xét vế trái

B=\(\frac{1}{n}.\frac{1}{n+4}\)=\(\frac{1}{n.\left(n+4\right)}\)

vì A=B --> điều phải chứng minh