\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
VM
12 tháng 11 2019
Ta có:
\(\Rightarrow\frac{x}{2008}=1\)
\(\Rightarrow x=1.2008\)
\(\Rightarrow x=2008\)
Vậy \(x=2008.\)
Chúc bạn học tốt!
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
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
TM
1
HT
16 tháng 9 2015
\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{120}=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
=> \(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}=1=\frac{2008}{2008}\)
=> x = 2008