Thực hiện phép tính:
A=3/4.8/9.15/16...9999/10000
B=(1-1/3)(1-1/6)(1-1/10)(1-1/15)...(1-1/780)
Giúp mk với mk đang cần gấp
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
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 10
A = \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\)...\(\dfrac{9999}{10000}\)
A = \(\dfrac{1.3.2.4..3.5......99.101}{2.2.3.3.4.4....100.100}\)
A = \(\dfrac{1.2.3..4.5.....99}{2.3.4.5.....99.100}\).\(\dfrac{3.4.5....100.101}{2.3.4.5...100}\)
A = \(\dfrac{1}{100}\).\(\dfrac{101}{2}\)
A = \(\dfrac{101}{200}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 10
2; B = (1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{8}\))...(1 - \(\dfrac{1}{n+1}\))
Xem lại đề bài.
TT
1
S
14 tháng 3 2017
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4...100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}.\frac{8}{9}....\frac{9999}{10000}=\frac{101}{200}\)
2 tháng 7 2015
A = \(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}=\frac{1\cdot3}{2.2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99\cdot101}{100\cdot100}=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)
B = ( 1- 1/4 )( 1-1/9) ...( 1-1/10000 ) = 3/4 . 8/9 .....9999/100000 ( tương tự A )
ND
1
HT
13 tháng 7 2015
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{2003}-1\right)\)
=\(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}.....\frac{-2002}{2003}\)
=\(\frac{1}{2003}\)
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
=\(\frac{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}\)
=\(\frac{101}{100.2}\)
=\(\frac{101}{200}\)
BT
1
KT
29 tháng 3 2018
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}......\frac{99.101}{100.100}\)
\(=\frac{1.2.3...99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)
\(=\frac{1}{100}.\frac{101}{2}\)
\(=\frac{101}{200}\)
3 tháng 9 2015
3/4.8/9.15/16......9999/10000
= 3.8.15.....9999/4.9.16......10000
=101/50
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 10
a; \(\dfrac{5}{6}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{20}\) + ... + \(\dfrac{5}{132}\)
= 5.(\(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + ..+ \(\dfrac{1}{132}\))
= 5.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{11.12}\))
= 5.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ...+ \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\))
= 5.(\(\dfrac{1}{2}\) - \(\dfrac{1}{12}\))
= 5.(\(\dfrac{6}{12}\) - \(\dfrac{1}{12}\))
= 5.\(\dfrac{5}{12}\)
= \(\dfrac{25}{12}\)
LT
2
LT
1
S
26 tháng 3 2017
\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{624}{625}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{24.26}{25.25}\)
\(=\frac{1.2.3....24}{2.3.4....25}\cdot\frac{3.4.5....26}{2.3.4....25}\)
\(=\frac{1}{25}\cdot\frac{26}{2}=\frac{26}{50}=\frac{13}{25}\)
\(\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{8}\right)\cdot\left(1+\frac{1}{15}\right)\cdot\cdot\cdot\cdot\left(1+\frac{1}{9999}\right)\)
\(=\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot\cdot\cdot\cdot\frac{10000}{9999}\)
\(=\frac{2.2}{1.3}\cdot\frac{3.3}{2.4}\cdot\frac{4.4}{3.5}\cdot\cdot\cdot\cdot\frac{100.100}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}\cdot\frac{2.3.4...100}{3.4.5...101}\)
\(=\frac{100}{1}\cdot\frac{2}{101}=\frac{200}{101}\)
KT
4 tháng 8 2018
mk chỉnh lại đề
\(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}....\frac{99}{100}\)
\(=\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}.\frac{5^2-1}{5^2}....\frac{10^2-1}{10^2}\)
\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}.....\frac{9.11}{10^2}\)
\(=\frac{2.3.4...9}{3.4.5...10}.\frac{4.5.6...11}{3.4.5...10}\)
\(=\frac{2}{10}.\frac{11}{3}=\frac{11}{15}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(A=\frac{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}{\left(2.3.4.....100\right).\left(2.3.4.....100\right)}\)
\(A=\frac{1.101}{2.100}=\frac{101}{200}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}.\frac{3.4.5.6.....101}{2.3.4.5.....100}\)
\(A=\frac{1}{100}.\frac{101}{2}\)
\(A=\frac{101}{200}\)