\(\Rightarrow\frac{x}{2013}-\left(\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\right)=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2013}-2\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2013}-2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2013}-\frac{3}{8}=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2013}=1\)

\(\Rightarrow x=2013\)

Ta có:

\(\frac{x}{2013}\)-\(\frac{1}{10}\)-\(\frac{1}{15}\)-\(\frac{1}{21}\)-...-\(\frac{1}{120}\)=\(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- (\(\frac{2}{20}\)+\(\frac{2}{30}\)+\(\frac{2}{42}\)+...+\(\frac{2}{240}\)) = \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{15.16}\)) = \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4}\)-\(\frac{1}{10}\)) = \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- 2.\(\frac{3}{10}\)= \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)= \(\frac{5}{8}\)+\(\frac{6}{10}\)= 1

=> \(x=2013\)

Vậy \(x=2013\)

2013 nha

khong hieu

\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)

\(\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)

\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)

\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}\)

\(\frac{x}{2008}=1=\frac{2008}{2008}\)

=> x = 2008

Vậy x = 2008

cảm ơn bạn mình cũng ra đáp án rồi

2008

Sai thì thôi nhé

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

Ta có:

\(\Rightarrow\frac{x}{2008}=1\)

\(\Rightarrow x=1.2008\)

\(\Rightarrow x=2008\)

Vậy \(x=2008.\)

Chúc bạn học tốt!