Question

chứng minh rằng n thuộc N,n khác 0 thì 

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

Answer 1

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

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

\(=\frac{1}{n}-\frac{1}{n+1}\) (đpcm)

Answer 2

Ta có : 

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

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

Tham khảo nha !!! 

Answer 3

ko chép cách giải 

tự làm ko được chép giải

Answer 4

\(\frac{1}{n\left(n+1\right)}\)

= \(\frac{n+1-n}{n\left(n+1\right)}\)

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

= \(\frac{1}{n}\)- \(\frac{1}{n+1}\)

 

Answer 5

Ta có : 1/n - 1/n+1 = n+1/n(n+1)- n/n(n+1)= n+1-n/n(n+1) = 1/n(n+1) (đpcm)