Question

\(3.Chứng\) Minh rằng

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

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

Answer 1

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

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

Tham khảo nha !!!! 

Answer 2

a, 

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

b,

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

Answer 3

a) Ta có:

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

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

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

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

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

b) Ta có:

\(\frac{1}{n}-\frac{1}{n+a}\)

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

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

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

\(=\frac{a}{n\left(n+a\right)}\left(đpcm\right)\)

Answer 4

Dễ ko

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

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