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!

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

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

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

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

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

=> \(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}=1=\frac{2008}{2008}\)

=> x = 2008