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\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}\)
\(\frac{x}{2008}=1=\frac{2008}{2008}\)
=> x = 2008
Vậy x = 2008
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
\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{120}=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
=> \(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}=1=\frac{2008}{2008}\)
=> x = 2008
b) Ta có \(\frac{3}{b}=\frac{a}{9}-\frac{1}{18}\) hay \(\frac{3}{b}=\frac{a.2}{18}-\frac{1}{18}=\frac{a.2-1}{18}\)
=> \(\left(a.2-1\right).b=3.18=54\)
Vì (a.2-1) là số lẻ => b là số chẵn
Ta có 54=1.54=27.2=3.18=9.6=(-1).(-54)=(-27).(-2)=(-3).(-18)=(-9).(-6)
Bạn tự tính nhé!
c) A=1-1\2+1\2-1\3+.......+1\99-1\100 câu c) so sánh ak!
A=1-1\100=99\100
1) Ta có: A=\(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\right)=\)
=\(\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\)
=\(\frac{1}{3}\left(1-\frac{1}{x+3}\right)=\frac{1}{3}.\frac{x+2}{x+3}=\frac{125}{376}\)
<=> \(\frac{x+2}{x+3}=\frac{375}{376}\)<=> 376(x+2)=375(x+3) <=> 376x+752=375x+1125 => X=373
