\(\frac{1}{120}-\frac{1}{30x33}-\frac{1}{33x36}-...-\frac{1}{117x120}\)
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NT
0
26 tháng 4 2019
A=1/8+1/24+1/48+1/80+1/120+1/168+1/224=>2A=2/8+2/24+2/48+2/80+2/120+2/168+2/224
2A=2/2*4+2/4*6+2/6*8+2/8*10+2/10*12+2/12*14+2/14*16
2A=1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12+1/12-1/14+1/14-1/16
2A=1/2-1/16
2A=7/16
A=7/16:2
A=7/32
US
0
US
0
TX
0
TT
2
7 tháng 7 2016
Ta có: \(B=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
\(\Rightarrow B=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)
\(\Rightarrow B=2.\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{4}-\frac{1}{16}\right)=2.\frac{3}{16}=\frac{3}{8}\)
Vậy \(B=\frac{3}{8}\)
nha m.n
VA
7 tháng 7 2016
\(B=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+.....+\frac{1}{120}\)
\(B=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{240}\right)\)
\(B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{15}-\frac{1}{16}\right)\)
\(B=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(B=2.\frac{3}{16}\)
\(B=\frac{3}{8}\)
Vậy \(B=\frac{3}{8}\)
NK
Tìm x biết:
\(\frac{x}{2013}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-.....-\frac{1}{120}=\frac{5}{8}\)
0
NT
4
4 tháng 3 2017
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 \(x=2013\)
cho mình hỏi dấu nhân có phải ko
\(=\frac{1}{120}-\left(\frac{1}{30.33}+\frac{1}{33.36}+...+\frac{1}{117.120}\right)\)
\(=\frac{1}{120}-\left[\frac{1}{3}.\left(\frac{3}{30.33}+\frac{3}{33.36}+...+\frac{3}{117.120}\right)\right]\)
\(=\frac{1}{120}-\left[\frac{1}{3}.\left(\frac{1}{30}-\frac{1}{33}+\frac{1}{33}-\frac{1}{36}+...+\frac{1}{117}-\frac{1}{120}\right)\right]\)