NV
1
K
Khách
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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
Q
16 tháng 4 2017
\(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-\dfrac{2}{20}-\dfrac{2}{30}-\dfrac{2}{42}-...-\dfrac{2}{240}=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\right)=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-2.\dfrac{3}{16}=\dfrac{5}{8}\)
\(\dfrac{x}{2008}-\dfrac{3}{8}=\dfrac{5}{8}\)
\(\dfrac{x}{2008}=\dfrac{5}{8}+\dfrac{3}{8}\)
\(\dfrac{x}{2008}=1=\dfrac{2008}{2008}\)
\(\Rightarrow x=2008\)
17 tháng 3 2017
b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=93-3=90\)
\(\Rightarrow x=90:2=45\)
KH
1
L
1
14 tháng 3 2016
hãy k nếu muốn
và nếu muốn thì hãy k mik bất cứ lúc nào
\(\dfrac{x-1}{2023}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)
=>\(\dfrac{x-1}{2023}-\left(\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{120}\right)=\dfrac{5}{8}\)
=>\(\dfrac{x-1}{2023}-2\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{240}\right)=\dfrac{5}{8}\)
=>\(\dfrac{x-1}{2023}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)
=>\(\dfrac{x-1}{2023}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)
=>\(\dfrac{x-1}{2023}-2\cdot\dfrac{3}{16}=\dfrac{5}{8}\)
=>\(\dfrac{x-1}{2023}=\dfrac{5}{8}+\dfrac{3}{8}=\dfrac{8}{8}=1\)
=>x-1=2023
=>x=2024