B=15/5.10+15/10.15+..........+15/100.105
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24 tháng 10 2017
a) \(D=\left(2\dfrac{2}{15}\times\dfrac{9}{17}\times\dfrac{3}{32}\right)\div\left(-\dfrac{3}{17}\right)\)
\(D=\dfrac{32}{15}\times\dfrac{9}{17}\times\dfrac{3}{32}\times\dfrac{-17}{3}\)
\(D=\dfrac{-3}{5}\)
b) \(\dfrac{1}{2}-\dfrac{1}{3\times7}-\dfrac{1}{7\times11}-\dfrac{1}{11\times15}-\dfrac{1}{15\times19}-\dfrac{1}{19\times23}-\dfrac{1}{23\times27}\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+\dfrac{1}{15\times19}+\dfrac{1}{19\times23}+\dfrac{1}{23\times25}\right)\)
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+\dfrac{4}{11\times15}+\dfrac{4}{15\times19}+\dfrac{4}{19\times23}+\dfrac{4}{23\times27}\right)\right]\)
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\right)\right]\)
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\right]\)
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{9-1}{27}\right)\right]\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\times\dfrac{8}{27}\)
\(=\dfrac{1}{2}-\dfrac{2}{27}\)
\(=.....\)
Đó đến đây bn tự lm nốt. Câu c bn lm tương tự.
Mình cho bn dạng này, nếu bn chưa biết (để lm câu c)
\(\dfrac{x}{y\left(y+x\right)}=\dfrac{x}{y}-\dfrac{x}{y+x}\)
Chúc bn học tốt
JL
4
8 tháng 8 2018
\(B=\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+...+\frac{5}{95\cdot100}\)
\(B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)
\(B=\frac{1}{5}-\frac{1}{100}\)
\(B=\frac{19}{100}\)
25 tháng 2 2022
\(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{395.400}\\ =\dfrac{1}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{395.400}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{395}-\dfrac{1}{400}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{400}\right)\\ =\dfrac{1}{5}.\dfrac{79}{400}\\ =\dfrac{79}{2000}\)
24 tháng 2 2022
\(\dfrac{2}{5.10}+\dfrac{2}{10.15}+...+\dfrac{2}{995.1000}\\ =2\left(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{995}-\dfrac{1}{1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{1000}\right)\)
\(=\dfrac{2}{5}.\dfrac{199}{1000}\\ =\dfrac{199}{2500}\)
31 tháng 7 2019
5/5.10 + 5/10.15 + ... + 5/45.50
= 1/5 - 1/10 + 1/10 - 1/15 + ... + 1/45 - 1/50
= 1/5 - 1/50
= 9/50
27 tháng 6 2017
\(A=\dfrac{3}{5\cdot10}+\dfrac{3}{10\cdot15}+...+\dfrac{3}{95\cdot100}\)
\(=\dfrac{3}{5}\left(\dfrac{5}{5\cdot10}+\dfrac{5}{10\cdot15}+...+\dfrac{5}{95\cdot100}\right)\)
\(=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{100}\right)\)\(=\dfrac{3}{5}\cdot\dfrac{19}{100}=\dfrac{57}{500}\)
27 tháng 6 2017
\(A=\dfrac{3}{5.10}+\dfrac{3}{10.15}+.....+\dfrac{3}{95.100}\)
\(A=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+.....+\dfrac{1}{95}-\dfrac{1}{100}\right)\)
\(A=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{5}.\dfrac{19}{100}=\dfrac{19}{500}\)
11 tháng 3 2017
C=1/5.10+1/10.15+...+1/95.100
= 5/5.10+5/10.15+...+5/95.100
= 1/5-1/10+1/10-1/15+...+1/95-1/100
= 1/5-1/100
= 19/100
\(B=\frac{15}{5.10}+\frac{15}{10.15}+....+\frac{15}{100.105}\)
\(B=3.\left(\frac{5}{5.10}+\frac{5}{10.15}+....+\frac{5}{100.105}\right)\)
\(B=3.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{105}\right)\)
\(B=3.\left(\frac{1}{5}-\frac{1}{105}\right)\)
\(B=3.\frac{4}{21}\)
\(B=\frac{4}{7}\)