(1/4-1)*(1/9-1)*...*(1/100-1)*(1/121-1)=?
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
VD
2
P5
16 tháng 5 2015
(1-1/4).(1-1/9).(1-1/16)....(1-1/100).(1-1/121)
=3/4.8/9.15/16...99/100.120/121
=(1.3/2.2).(2.4/3.3).(3.5/4.4)....(9.11/10.10).(10.12/11.11)
=1.3.2.4.3.5...9.11.10.12/2.2.3.3.4.4...10.10.11.11
=(2.3.4...9.10.11).(3.4.5...10.12)/(2.3.4...9.10.11).(2.3.4....10.11)
=12/2.11
=6/11
ND
2
NA
Nguyễn An Ninh
VIP
22 tháng 4 2023
(1-1/4).(1-1/9).(1-1/16)....(1-1/100).(1-1/121)
=3/4.8/9.15/16...99/100.120/121
=(1.3/2.2).(2.4/3.3).(3.5/4.4)....(9.11/10.10).(10.12/11.11)
=1.3.2.4.3.5...9.11.10.12/2.2.3.3.4.4...10.10.11.11
=(2.3.4...9.10.11).(3.4.5...10.12)/(2.3.4...9.10.11).(2.3.4....10.11)
=12/2.11
=6/11
nha
PT
0
26 tháng 4 2015
\(\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...\times\frac{99}{100}\times\frac{120}{121}=\frac{3\times8\times15\times...\times99\times120}{4\times9\times16\times...\times100\times121}\)
\(\frac{\left(1\times3\right)\times\left(2\times4\right)\times\left(3\times5\right)\times...\times\left(9\times11\right)\times\left(10\times12\right)}{\left(2\times2\right)\times\left(3\times3\right)\times\left(4\times4\right)\times...\times\left(10\times10\right)\times\left(11\times11\right)}=\frac{\left(1\times2\times3\times...\times10\right)\times\left(3\times4\times5\times...\times12\right)}{\left(2\times3\times...\times11\right)\times\left(2\times3\times...\times11\right)}=\frac{12}{11\times2}=\frac{6}{11}\)
26 tháng 4 2015
(1-1/4).(1-1/9).(1-1/16)....(1-1/100).(1-1/121)
=3/4.8/9.15/16...99/100.120/121
=(1.3/2.2).(2.4/3.3).(3.5/4.4)....(9.11/10.10).(10.12/11.11)
=1.3.2.4.3.5...9.11.10.12/2.2.3.3.4.4...10.10.11.11
=(2.3.4...9.10.11).(3.4.5...10.12)/(2.3.4...9.10.11).(2.3.4....10.11)
=12/2.11
=6/11
L
1
25 tháng 7 2020
Bạn ấn vào chữ "Câu hỏi tương tự" để tham khảo nhé
( nếu ko bt thì copy link này : https://olm.vn/hoi-dap/tim-kiem?id=259990292185&id_subject=1&q=+++++++++++[+1-1/4+]+x+[1-1/9+]x+[+1-+1/16+]+x..........x+[+1-1/100+]+x+[+1-+1/121+]++++++++++ )
DT
2
HT
13 tháng 7 2015
\(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{100}\right)\left(1-\frac{1}{121}\right)\)
=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{99}{100}.\frac{120}{121}\)
=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{3.3}.....\frac{9.11}{10.10}.\frac{10.12}{11.11}\)
=\(\frac{\left(1.2.3.....9.10\right)\left(3.4.5.....11.12\right)}{\left(2.3.4.5.....10.11\right)\left(2.3.4.5....10.11\right)}\)
=\(\frac{12}{11.2}\)
=\(\frac{6}{11}\)
US
1
KT
24 tháng 7 2018
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}+\frac{1}{121}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}+\frac{1}{11^2}\)
Ta có: \(\frac{1}{2^2}>\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3^2}>\frac{1}{3}-\frac{1}{4}\)
\(\frac{1}{4^2}>\frac{1}{4}-\frac{1}{5}\)
................................
\(\frac{1}{10^2}>\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11^2}>\frac{1}{11}-\frac{1}{12}\)
Cộng theo vế ta được:
\(A>\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
Vậy \(A>\frac{5}{12}\)
\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).....\left(\frac{1}{101}-1\right)\)
\(=\frac{3}{4}.\frac{8}{9}.....\frac{120}{121}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{10.12}{11.11}\)
\(=\frac{\left(1.3\right).\left(2.4\right).....\left(10.12\right)}{\left(2.2\right).\left(3.3\right).....\left(11.11\right)}\)
\(=\frac{\left(1.2.3.....10\right).\left(3.4.5.....12\right)}{\left(2.3.4.....11\right).\left(2.3.4.....11\right)}\)
\(=\frac{1.12}{11.2}=\frac{6}{11}\)