\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}..........\frac{9999}{10000}\)
K
Khách
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MT
1 tháng 7 2015
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100.100}\)
=\(\frac{1.3.2.4.3.5....999.101}{2.2.3.3.4.4....100.100}=\frac{1.101}{2.100}=\frac{101}{200}\)
NT
0
P
1
BM
9 tháng 4 2016
Ta có :
\(A<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.............\frac{10000}{10001}=M\)
=> A.A < A.M = \(\frac{1}{10001}\)
=> A2 < \(\frac{1}{10000}=\left(\frac{1}{100}\right)^2\)
=> A < \(\frac{1}{100}\)
k nha bạn
NT
2 tháng 4 2015
\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{10000}\)
\(=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot...\cdot\frac{99\cdot101}{100^2}\)
\(=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(99\cdot101\right)}{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}\)
\(=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(=2\cdot101=202\)
PH
4
N
17 tháng 10 2018
\(C=\frac{3.8.15....80.99}{4.9.16.81.100}\)
\(=\frac{1.3.2.4.3.5...8.10.9.11}{2.2.3.3.4.4...9.9.10.10}\)
\(=\frac{\left(1.2.3....9\right).\left(3.4.5...10.11\right)}{\left(2.3.4.5...10\right).\left(2.3.4...10\right)}\)
\(=\frac{11}{10}\)
17 tháng 10 2018
trả lời
c=\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot....\cdot\frac{99}{100}\)
C=\(\frac{3.8.15....99}{4.9.16.100}\)
C=\(\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4....10.10}\)
C=\(\frac{\left(1.2.....9\right)}{2.3....10}.\left(\frac{3.4....11}{2.3...10}\right)\)
C=\(\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)
30 tháng 4 2019
\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{99}{100}\)
\(=\frac{3.8.15.24....99}{4.9.16.25....100}\)
\(=\frac{1.3.2.4.3.5.4.6....9.11}{2.2.3.3.4.4.5.5....10.10}\)
\(=\frac{1.2.3.4....9}{2.3.4.5....10}.\frac{3.4.5.6....11}{2.3.4.5....10}\)
\(=\frac{1}{10}.\frac{11}{2}\)
\(=\frac{11}{20}\)
Study well ! >_<
KN
1 tháng 5 2019
\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{99}{100}\)
\(\Rightarrow M=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{9.11}{10.10}\)
\(\Rightarrow M=\frac{1.3.2.4.3.5...9.11}{2.2.3.3.4.4...10.10}\)
\(\Rightarrow M=\frac{\left(1.2.3...9\right)\left(3.4.5...11\right)}{\left(2.3.4...10\right)\left(2.3.4...10\right)}\)
\(\Rightarrow M=\frac{11}{10.2}\)
\(\Rightarrow M=\frac{11}{20}\)
18 tháng 2 2017
\(A=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times...\times\frac{899}{900}\)
\(=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times...\times\frac{29.31}{30.30}\)
\(=\frac{\left(1\times2\times3\times...\times29\right)\left(3\times4\times5\times...\times31\right)}{\left(2\times3\times4\times...\times30\right)\left(2\times3\times4\times...\times30\right)}\)
\(=\frac{1\times2\times3\times...\times29}{2\times3\times4\times...\times30}.\frac{3\times4\times5\times...\times31}{2\times3\times4\times...\times30}\)
\(=\frac{1}{30}.\frac{31}{2}\)
\(=\frac{31}{60}\)
NT
19 tháng 2 2017
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{899}{900}\\ =\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{29.31}{30.30}\\ =\frac{1.2.3.4....29}{2.3.4...30}.\frac{3.4.5...31}{2.3.4...30}\\ =\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)
.
5 tháng 7 2016
Ta có:\(B=\frac{1.3}{2.2}x\frac{2.4}{3.3}x\frac{3.5}{4.4}x...x\frac{10.12}{11.11}\)
\(=\frac{1.2.3.4.5.6.7.8.9.10}{2.3.4.5.6.7.8.9.10.11}x\frac{3.4.5.6.7.8.9.10.11.12}{2.3.4.5.6.7.8.9.10.11}\)
\(=\frac{1}{11}x6\)
\(=\frac{6}{11}\)
Ai thấy đúng k nha.
TD
24 tháng 3 2017
Đặt C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)\(\left(C>0\right)\)
Và D = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{10000}{10001}\)\(\left(D>0\right)\)
Ta có :
C .D = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10000}{10001}\)\(=\frac{1}{10001}\)\(\left(1\right)\)
Mặt khác :
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(.....\)
\(\frac{9999}{10000}< \frac{10000}{10001}\)
Nhân tất cả vế theo vế - - - > C < D - - - > C2 < C . D \(\left(2\right)\)
\(\left(1\right),\left(2\right)\)- - - >C2 < \(\frac{1}{10001}\)- - - > C < căn \(\frac{1}{10001}\)< căn \(\frac{1}{10000}\)= \(\frac{1}{100}\)( đpcm )
