Question

[★Čүċℓøρş★]

\(3.Chứng\) 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

[Arima Kousei]

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

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

Tham khảo nha !!!! 

Answer 2

[Nguyễn Phương Uyên]

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,

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