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

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

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

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

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

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

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

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

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

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

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

=> 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

Ko đánh đc phân số nên cho kết quả lun nha: 2013

1) 

a) \(-\frac{8}{15}< \frac{x}{45}< -\frac{2}{5}\)

Lại có: \(-\frac{8}{15}=\frac{-24}{45};-\frac{2}{5}=\frac{-18}{45}\)

=> \(-\frac{24}{45}< \frac{x}{45}< -\frac{18}{45}\)

=> -24 < x < - 18 

=> x \(\in\){ - 23; -22; -21; -20 ; -19 } ( thử lại thỏa mãn )

b) \(x=\frac{-4}{3}+\frac{-7}{5}=-\frac{4.5}{3.5}+\frac{-7.3}{5.3}=-\frac{41}{15}\)

c) \(\frac{83}{x}=\frac{13}{4}+\frac{9}{10}=\frac{83}{20}\)

=> x = 20 ( thử lại thỏa mãn) 

d) \(x=\frac{10}{8}+\frac{-24}{48}+\frac{105}{-120}=-\frac{1}{8}\)

e) \(\left|x-\frac{1}{2}\right|=\left|-\frac{2}{7}\right|+\frac{5}{4}\)

\(\left|x-\frac{1}{2}\right|=\frac{2}{7}+\frac{5}{4}\)

\(\left|x-\frac{1}{2}\right|=\frac{43}{28}\)

TH1: \(x-\frac{1}{2}=\frac{43}{28}\)

         \(x=\frac{57}{28}\)

TH2: \(x-\frac{1}{2}=-\frac{43}{28}\)

      \(x=-\frac{29}{28}\)

1)

a)

\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)

\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)

\(\frac{-20}{5}< x< \frac{-3}{10}\)

\(\frac{-40}{10}< x< \frac{-3}{10}\)

\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)

\(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)

\(\frac{25}{3}< x< \frac{-4}{7}.\frac{1}{1}\)

\(\frac{-25}{3}< x< \frac{-4}{7}\)

\(\frac{-175}{21}< x< \frac{-12}{21}\)

\(\Rightarrow Z\in\left\{-13;-14;-15;-16;...;-174\right\}\)

2)

a)

\(\left(\frac{9}{10}-\frac{15}{16}\right).\left(\frac{5}{12}-\frac{11}{15}-\frac{7}{20}\right)\)

\(=\left(\frac{72}{80}-\frac{75}{80}\right).\left(\frac{25}{60}-\frac{44}{60}-\frac{21}{60}\right)\)

\(=\frac{-3}{80}.\frac{-40}{60}\)

\(=\frac{-1}{-2}.\frac{-1}{-20}\)

\(=\frac{1}{40}\)