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LL
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TL
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SA
18 tháng 11 2018 lúc 6:45

\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\)

\(A=\dfrac{1}{6}.\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+...+\dfrac{6}{50.56}\right)\)

\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+...+\dfrac{1}{50}-\dfrac{1}{56}\right)\)

\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)\)

\(A=\dfrac{1}{6}.\left(\dfrac{7}{56}-\dfrac{1}{56}\right)\)

\(A=\dfrac{1}{6}.\dfrac{6}{56}\)

\(A=\dfrac{1}{1}.\dfrac{1}{56}\)

\(A=\dfrac{1}{56}\)

\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)

\(B=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+\dfrac{10}{34.44}+\dfrac{10}{44.54}+\dfrac{10}{54.64}\right)\)

\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{5}{60}-\dfrac{2}{60}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)

\(B=5.\dfrac{3}{60}-\left(\dfrac{4}{24}-\dfrac{4}{64}\right)\)

\(B=5.\dfrac{1}{20}-\left(\dfrac{1}{6}-\dfrac{1}{16}\right)\)

\(B=\dfrac{5}{20}-\left(\dfrac{8}{48}-\dfrac{3}{48}\right)\)

\(B=\dfrac{1}{4}-\dfrac{5}{48}\)

\(B=\dfrac{12}{48}-\dfrac{5}{48}\)

\(B=\dfrac{7}{48}\)

\(\dfrac{A}{B}=\dfrac{1}{56}:\dfrac{7}{48}\)

\(\dfrac{A}{B}=\dfrac{1}{56}.\dfrac{48}{7}\)

\(\dfrac{A}{B}=\dfrac{1}{7}.\dfrac{6}{7}\)

\(\dfrac{A}{B}=\dfrac{6}{49}=\dfrac{48}{392}< \dfrac{49}{392}=\dfrac{1}{8}\)

\(\dfrac{A}{B}< \dfrac{1}{8}\)

Vậy \(\dfrac{A}{B}< \dfrac{1}{8}\)

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VD
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VH
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EC
23 tháng 3 2017 lúc 19:57

b)\(\dfrac{1}{7}B=\dfrac{1}{10.18}+\dfrac{1}{18.26}+\dfrac{1}{26.34}+...+\dfrac{1}{802.810}\)

\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{8}{10.18}+\dfrac{8}{18.26}+\dfrac{8}{26.34}+...+\dfrac{8}{802.810}\right)\)

\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{34}+...+\dfrac{1}{802}-\dfrac{1}{810}\right)\)

\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{810}\right)\)

\(\dfrac{1}{7}B=\dfrac{1}{8}.\dfrac{8}{81}\)

\(\dfrac{1}{7}B=\dfrac{1.8}{8.81}\)

\(\dfrac{1}{7}B=\dfrac{1}{81}\)

\(B=\dfrac{1}{81}:\dfrac{1}{7}\)

\(B=\dfrac{7}{81}\)

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PT
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CD
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ND
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DG
22 tháng 9 2023 lúc 22:39

1+1

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DG
22 tháng 9 2023 lúc 22:40

2/3+3/2

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TL
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H24
8 tháng 6 2017 lúc 7:34

b,

\(B=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)

\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)

\(\Rightarrow B=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)

\(\Rightarrow B=\frac{1}{1999.2000}-\frac{1998}{1999}\)

\(\Rightarrow B=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)

\(\Rightarrow B=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)

\(\Rightarrow B=\frac{-1997}{1999}-\frac{1}{2000}\)

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KK
8 tháng 6 2017 lúc 14:09

A= (1-\(\dfrac{1}{2}\) ) . ( 1- \(\dfrac{1}{3}\) ) . ( 1 - \(\dfrac{1}{4}\)) ...........( 1- \(\dfrac{1}{n+1}\))

= \(\dfrac{1}{2}\). \(\dfrac{2}{3}\) . \(\dfrac{3}{4}\) ......\(\dfrac{1}{n+1}\)

= 1

Mik nghĩ chắc là thê này đó Khánh Linh

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H24
8 tháng 6 2017 lúc 15:18

a)

Có:

\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{n+1}\right)\)

\(\Leftrightarrow A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{n}{n+1}\)

Rút gọn biểu thức A, ta được:

\(A=\dfrac{1}{n-1}\)

Chúc bạn học tốt!ok

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DT
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LN
7 tháng 3 2017 lúc 8:22

22 là thế nào đấy bạn?

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