TH
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TT
2
7 tháng 7 2016
Ta có: \(B=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
\(\Rightarrow B=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)
\(\Rightarrow B=2.\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{4}-\frac{1}{16}\right)=2.\frac{3}{16}=\frac{3}{8}\)
Vậy \(B=\frac{3}{8}\)
nha m.n
VA
7 tháng 7 2016
\(B=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+.....+\frac{1}{120}\)
\(B=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{240}\right)\)
\(B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{15}-\frac{1}{16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(B=2.\frac{3}{16}\)
\(B=\frac{3}{8}\)
Vậy \(B=\frac{3}{8}\)
NH
0
RK
2
Q
16 tháng 3 2018
Có: \(N=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
\(=>N=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=>N=\frac{2}{4\cdot5}+\frac{2}{5\cdot6}+\frac{2}{6\cdot7}+...+\frac{2}{15\cdot16}\)
\(=>N=\left(\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+...+\frac{2}{15}-\frac{2}{16}\right)\)
\(=>N=\frac{2}{4}-\frac{2}{16}\)
\(=>N=\frac{1}{2}-\frac{1}{8}\)
\(=>N=\frac{8-2}{16}=\frac{6}{16}=\frac{3}{8}\)
Vậy \(N=\frac{3}{8}\)
16 tháng 3 2018
Ta có :
\(N=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(N=2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(N=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(N=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(N=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(N=\frac{1}{2}-\frac{1}{8}\)
\(N=\frac{3}{8}\)
Vậy \(N=\frac{3}{8}\)
Chúc bạn học tốt ~
T
2
6 tháng 4 2019
\(=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{15.16}\)
\(=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
=\(\frac{1}{8}\)
Tích cho mình nhé cảm ơn
KN
7 tháng 4 2019
\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)
\(=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=2\left(\frac{4}{16}-\frac{1}{16}\right)\)
\(=2\times\frac{3}{16}\)
\(=\frac{6}{16}=\frac{3}{8}\)
25 tháng 6 2016
\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}\)
\(\frac{x}{2008}=1=\frac{2008}{2008}\)
=> x = 2008
Vậy x = 2008
P
1
PN
5 tháng 5 2016
Đặt biểu thức đã cho là A
A= -( 1/15 + 1/21 + ...+1/120 - 1/10)
= -2(1/30 + 1/42 +...+ 1/240 - 1/20)
= -2( 1/5-1/6 + 1/6-1/7 + ...+ 1/15-1/16 - 1/4 + 1/5)
= -2 ( 1/5 -1/16 - 1/4 +1/5)
= -2( 2/5 -5/16)
= -2 . 7/80
= -7/40
Vậy A= -7/40
NH
2
NH
2
LT
22 tháng 9 2019
2008x−101−151−...−1201=85
\frac{x}{2008}-2.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)=\frac{5}{8}2008x−2.(4.51+5.61+...+15.161)=85
\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}2008x−2.(41−51+51−61+....+151−161)=85
\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}2008x−2.(41−161)=85
\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}2008x−83=85
=> \frac{x}{2008}=\frac{5}{8}+\frac{3}{8}=1=\frac{2008}{2008}2008x=85+83=1=20082008
=> x = 2008
\(=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{420}\)
\(=\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{20.21}\)
\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{21}\right)\)
\(=2\times\frac{17}{84}\)
\(=\frac{17}{72}\)