(1 - 1/4) \(\times\)(1 - 1/7) \(\times\)( 1 - 1/10 ) \(\times\).............. \(\times\)( 1 - 1/97 ) \(\times\)( 1 - 1/100 )
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TP
1
NG
1 tháng 6 2020
A=\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)+...+\(\frac{1}{99.100}\)
⇒A=\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)+...+\(\frac{1}{99}\)-\(\frac{1}{100}\)
⇒A=\(\frac{1}{10}\)-\(\frac{1}{100}\)
⇒A=\(\frac{9}{100}\)
Vậy A=\(\frac{9}{100}\)
B=\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+...+\(\frac{1}{97.99}\)
=\(\frac{1}{2}\).\((1-\frac{1}{3})\)+\(\frac{1}{2}.(\frac{1}{3}-\frac{1}{5})\)+...+\(\frac{1}{2}.(\frac{1}{97}-\frac{1}{99})\)
=\(\frac{1}{2}.(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99})\)
=\(\frac{1}{2}.\frac{98}{99}\)
=\(\frac{49}{99}\)
Vậy B=\(\frac{49}{99}\)
19 tháng 11 2018
Bài 3: a) Xét A=(1+1/2+1/3+....+1/98).2.3.4.5.....98
=(1+1/2+1/3+....+1/98).(9.11).2.3.4.....98
=(1+1/2+1/3+....+1/98).99.2.3.4....98⋮99
(đpcm)
BT
1
28 tháng 4 2017
\(\dfrac{1}{2\cdot5}+\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot7}+\dfrac{1}{4\cdot7}+...+\dfrac{1}{9\cdot19}+\dfrac{1}{10\cdot19}=\dfrac{3+2}{2.3.5}+\dfrac{4+3}{3\cdot4\cdot7}+...+\dfrac{10+9}{9\cdot10\cdot19}=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9.10}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\)
HH
10 tháng 6 2018
Bài 2:
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{101}\)
\(\Leftrightarrow A=\dfrac{100}{101}\)
Vậy ...
\(B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+\dfrac{1}{13.16}\)
\(\Leftrightarrow B=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}.\dfrac{15}{16}\)
\(\Leftrightarrow B=\dfrac{5}{16}\)
Vậy ...
10 tháng 6 2018
Bài 1:
B=\(\dfrac{\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\right)}{\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}\right)}\)
\(=\dfrac{\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}\right)}{\left(1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}\right)}\)
\(=\dfrac{\left(\dfrac{2^4+2^3+2^2+2+1}{2^4}\right)}{\left(2^4-2^3+2^2-2+1\right)}\)
\(=\dfrac{\left(2^3+2\right)\left(2+1\right)+1}{2^4}.\dfrac{2^4}{\left(2^3+2\right)\left(2-1\right)}\)
\(=\dfrac{2\left(2^2+1\right)\left(2+1\right)+1}{2\left(2^2+1\right)\left(2-1\right)+1}\)
\(=\dfrac{2.5.3+1}{2.5.1+1}\)
\(=\dfrac{31}{11}\)
\(=2,\left(81\right)\)
27 tháng 9 2020
a,Đặt \(A=\frac{1}{1\times4}+\frac{1}{4\times7}+...+\frac{1}{97\times100}\)
\(\Rightarrow3A=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{97\times100}\)
\(\Rightarrow3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{300}\)
b, \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}=\frac{1\times2\times...\times99}{2\times3\times...\times1000}=\frac{1}{100}\)
27 tháng 9 2020
c, \(\frac{3}{4}\times\frac{8}{9}\times...\times\frac{99}{100}=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times...\times\frac{9.11}{10.10}=\frac{1.2.....9}{2.3.....10}\times\frac{3.4.....11}{2.3.....10}=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\) (dấu . là dấu nhân)
Q
12 tháng 4 2017
a, \(A=\dfrac{1}{3}.\dfrac{-6}{-3}.\dfrac{-9}{10}.\dfrac{-13}{36}\)
\(A=\dfrac{1.\left(-6\right).\left(-9\right).\left(-13\right)}{3.13.10.36}\)
\(A=\dfrac{-1}{10.2}\)
\(A=\dfrac{-1}{20}\)
b, \(B=\dfrac{-1}{3}.\dfrac{-15}{17}.\dfrac{34}{45}\)
\(B=\dfrac{\left(-1\right).\left(-15\right).34}{3.17.45}\)
\(B=\dfrac{2}{3.3}\)
\(B=\dfrac{2}{9}\)
c, \(C=\dfrac{1}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{6}{5}+\dfrac{2}{3}\)
\(C=\dfrac{1}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\)
\(C=\dfrac{1}{3}.2+\dfrac{2}{3}\)
\(C=\dfrac{2}{3}+\dfrac{2}{3}\)
\(C=\dfrac{4}{3}\)
d, \(D=\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}-\dfrac{40}{57}\)
\(D=\dfrac{4}{19}.\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)
\(D=\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)
\(D=\dfrac{-17}{57}-\dfrac{40}{57}\)
\(D=\dfrac{-57}{57}=-1\)
e, \(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{14}.\dfrac{1}{13}-\dfrac{1}{7}\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{14}.\dfrac{1}{13}+\dfrac{1}{7}\right)\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{182}+\dfrac{1}{7}\right)\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{27}{182}\)
\(E=\dfrac{27}{182}-\dfrac{27}{182}\)
\(E=0\)