K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

10 tháng 5 2017

Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)

ai giúp mk ik

mk đg cần gấp,còn nhìu đề chx lm

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(\Rightarrowđpcm\)

10 tháng 4 2018

free ire

20 tháng 3 2016

Số chia rút gọn thành 1/51+1/52+...+1/99+1/100

=> biểu thức bằng 1

9 tháng 5 2017

Ta có :\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\)\(\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\)\(+...+\left(1-\frac{1}{100}\right)\)

=\(\left(1+1+1+....+1\right)\)\(-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=             \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=  \(100-1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=\(100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)= vế trên (đpcm)

9 tháng 5 2017

\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)

2 tháng 4 2023

1+1=3 :)))

27 tháng 4 2018

\(VT=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{101}+\frac{1}{102}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}+\frac{1}{102}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{51}\)

\(=\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{102}\)

\(=VP\)

22 tháng 1 2017

\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\left(1+1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\) ( ĐPCM )