Question

Chứng minh rằng với \(n\in N\), \(n\ne0\) thì:

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

Answer 1

bien doi ve phai;

qui dong mau so ta co(( n+1)-n )/n(n+1) = 1/n(n+1)= ve trai (dpcm)

de hon an com suon

;

Answer 2

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

Answer 3

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

Answer 4

ta có: 1/n - 1/n+1 = (n+1)1/n(n+1) - 1n/n(n+1) = n+1-n/ n(n+1) = 1/n(n+1)

vì 1/n(n+1)= 1/n(n+1) nên 1/n - 1/n+1 = 1/ n(n+1)   ( đccm )

Answer 5

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