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5 tháng 8 2016
Ta có:
1/4 = 1/3( 1-1/4)
1/28 = 1/3( 1/4 - 1/7)
1/70 = 1/3( 1/7 - 1/10)
..............................
1/10300 = 1/3( 1/100 - 1/103)
Cộng vế với vế ta có:
S = 1/4+1/28+1/70+1/130+...+1/10300 = 1/3( 1-1/103)
S = 34/103
19 tháng 4 2020
\(3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
\(3M=\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{100-97}{97.100}\)
\(3M=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(3M=1-\frac{1}{100}\)
\(3M=\frac{99}{100}\)
\(M=\frac{33}{100}\)
MV
1
2 tháng 7 2017
\(4-\dfrac{1}{4}-\dfrac{1}{28}-\dfrac{1}{70}-.....-\dfrac{1}{2002.2005}\)
\(=4-\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{2002.2005}\right)\)
\(=4-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)\(=4-\left(1-\dfrac{1}{2005}\right)\)
\(=4-1+\dfrac{1}{2005}=3+\dfrac{1}{2005}=\dfrac{6016}{2005}\)
12 tháng 8 2016
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}\)
\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{16}\right)\)
\(=\frac{1}{3}.\frac{15}{16}=\frac{5}{16}\)
PL
0
DT
1
PH
1
GK
30 tháng 3 2018
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{9700}=\frac{0,33x}{2009}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{100}{100}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{2009}\)
\(\Rightarrow2009.99=100.0,33x\)
\(\Rightarrow2009.99=33x\)
\(\Rightarrow2009.99:33=x\)
\(\Rightarrow2009.3=x\)
\(\Rightarrow6027=x\)
Vậy \(x=6027\)(MK KO CHẮC NÓ ĐÚNG NHÉ )
Đặt :
\(M=\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+................+\dfrac{1}{550}\)
\(\Rightarrow M=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+..................+\dfrac{1}{22.25}\)
\(\Rightarrow3M=\dfrac{3}{1.4}+\dfrac{3}{4.7}+.................+\dfrac{3}{22.25}\)
\(\Rightarrow3M=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.........+\dfrac{1}{22}-\dfrac{1}{25}\)
\(\Rightarrow3M=1-\dfrac{1}{25}\)
\(\Rightarrow3M=\dfrac{24}{25}\)
\(\Rightarrow M=\dfrac{24}{75}\)
Đặt:
\(A=\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+.....+\dfrac{1}{550}\)
\(A=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{22.25}\)
\(A=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{22}-\dfrac{1}{25}\right)\)
\(A=\dfrac{1}{3}\left(1-\dfrac{1}{25}\right)=\dfrac{1}{3}.\dfrac{24}{25}=\dfrac{25}{72}\)