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PT
16 tháng 4 2016
\(B=2.\left(\frac{1}{30}+\frac{1}{42}+...+\frac{1}{380}\right)\)
\(B=2.\left(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{19.20}\right)\)
\(B=2.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(B=2.\left(\frac{1}{5}-\frac{1}{20}\right)\)
\(B=2.\frac{3}{20}\)
\(B=\frac{3}{10}\)
Đáp án này là chính xác nhất đó, mình thề. k cho mình nhé
KT
6 tháng 4 2018
1/15+1/21+1/28+...+1/190
=2/30+2/42+2/56+...+2/380
=2/5.6+2/6.7+2/7.8+...+2/19.20
=2.(1/5.6+1/6.7+1/7.8+...+1/19.20)
=2.(1/5-1/6+1/6-1/7+1/7-1/8 +...+1/19-1/20)
=2.(1/5-1/20)
=2.3/20
=6/20=3/10
S
3 tháng 5 2017
\(C=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)
\(C=\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{380}\)
\(C=2\left(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{19.20}\right)\)
\(C=2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(C=2\left(\frac{1}{5}-\frac{1}{20}\right)\)
\(C=2\cdot\frac{3}{20}=\frac{3}{10}\)
NH
1
21 tháng 4 2016
- 2.s= 1/30+1/42+1/56+...+ 1/380
2.S= 1/ 5.6 =1/ 6.7 +1/ 7.8 +...+1/ 19.20
2.S= 1/5-1/20
2S= 3/20
15 tháng 6 2019
\(A=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)
\(A=\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{380}\) ( nhân cả tử và mẫu với 2 )
\(A=\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{19.20}=2\left(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{19.20}\right)\)
A = \(2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\right)=2\left(\frac{1}{5}-\frac{1}{20}\right)=2.\frac{3}{20}=\frac{3}{10}\)
B = \(\frac{12}{84}+\frac{12}{210}+\frac{12}{390}+...+\frac{12}{2100}\)
\(B=\frac{4}{28}+\frac{4}{70}+\frac{4}{130}+...+\frac{4}{700}\) ( chia cả tử và mẫu của mỗi phân số cho 3 )
B = \(\frac{4}{4.7}+\frac{4}{7.10}+\frac{4}{10.13}+...+\frac{4}{25.28}=\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)
B = \(\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{4}{3}.\frac{6}{28}=\frac{2}{3}\)
TH
1
17 tháng 4 2018
\(F=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)
\(\Rightarrow F=\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{380}\)
\(\Rightarrow F=\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{19.20}\)
\(\Rightarrow F=2.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(\Rightarrow F=2.\left(\frac{1}{5}-\frac{1}{20}\right)\)
\(\Rightarrow F=2.\frac{3}{20}\)
\(\Rightarrow F=\frac{3}{10}\)
21 tháng 7 2017
+) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^{10}}=\dfrac{2^{10}-1}{2^{10}}\)
Vậy \(A=\dfrac{2^{10}-1}{2^{10}}\)
+) \(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)
\(\Rightarrow\dfrac{1}{2}F=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{380}\)
\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=\dfrac{1}{5}-\dfrac{1}{20}=\dfrac{3}{20}\Rightarrow F=\dfrac{3}{20}:\dfrac{1}{2}=\dfrac{3}{10}\)
Vậy \(F=\dfrac{3}{10}\)
+) \(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)
\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}=\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{25.28}\)
\(=\dfrac{4}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{25.28}\right)\)
\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)
Vậy \(G=\dfrac{2}{7}\)
21 tháng 7 2017
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
\(A=1-\dfrac{1}{2^{10}}=\dfrac{1024-1}{1024}=\dfrac{1023}{1024}\)
\(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)
\(=\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{380}\)
\(=\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{19.20}\)
\(=2\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{20}\right)=2.\dfrac{3}{20}=\dfrac{3}{10}\)
\(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)
\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}\)
\(=\dfrac{4}{4.7}+\dfrac{4}{7.10}+\dfrac{4}{10.13}+...+\dfrac{4}{25.28}\)
\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)
B=2.(130+142+...+1380)B=2.(130+142+...+1380)
B=2.(15.6+16.7+...+119.20)B=2.(15.6+16.7+...+119.20)
B=2.(15−16+16−17+...+119−120)B=2.(15−16+16−17+...+119−120)
B=2.(15−120)B=2.(15−120)
B=2.320B=2.320
B=310
\(B=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)
\(B=\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{380}\)
\(B=2\left(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{19.20}\right)\)
\(B=2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(B=2\left(\frac{1}{5}-\frac{1}{20}\right)\)
\(B=2\cdot\frac{3}{20}=\frac{3}{10}\)