Tính \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\div\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\div\frac{919191}{808080}\)
K
Khách
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Những câu hỏi liên quan
21 tháng 2 2016
Gọi \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)
\(B=1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\)
Từ đề bài ta có
\(D=182\left[\frac{A}{2A}:\frac{4B}{B}\right]:\frac{919191}{808080}\)
\(D=182\times\left(\frac{1}{2}:4\right):\frac{91}{80}\)
\(D=182\times\frac{1}{8}\times\frac{80}{91}\)
\(D=\frac{91\times2\times1\times8\times10}{8\times91}=20\)
cho tui nha
HP
21 tháng 2 2016
Ta có:\(D=182\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(D=182\left[\frac{1\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{2}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(D=182\left[\frac{1}{2}:4\right]:\frac{919191}{808080}=182\left[\frac{1}{2}.\frac{1}{4}\right]:\frac{919191}{808080}=182.\frac{1}{8}:\frac{919191}{808080}=\frac{182}{8}:\frac{919191}{808080}\)Mà \(\frac{919191}{808080}=\frac{919191:10101}{808080:10101}=\frac{91}{80}\)
\(\Rightarrow D=\frac{182}{8}:\frac{91}{80}=\frac{182}{8}.\frac{80}{91}=\frac{182.80}{8.91}=\frac{91.2.8.10}{8.91}=2.10=20\)
Vậy D=20
7 tháng 8 2015
= \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}+\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\right):\frac{91}{80}\)
= \(\frac{1}{2}:4:\frac{91}{80}=\frac{10}{91}\)
F
29 tháng 6 2019
Bài giải
\(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)
\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)
MG
11 tháng 4 2018
=182.\(\orbr{\begin{cases}1.\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\\2.\left(\frac{1}{2}+\frac{1}{9}+\frac{1}{27}\right)\end{cases}}:\frac{4.\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1.\left(\frac{1}{3}+\frac{1}{49}-\frac{1}{343}\right)}:\frac{91}{80} \)
=.\(182.\left(\frac{1}{2}:\frac{4}{1}\right).\frac{91}{80}\)
=\(182.\frac{1}{8}.\frac{91}{80}\)
=.\(182.\frac{91}{640}\)
=\(\frac{8281}{320}\)
B
11 tháng 4 2018
\(=182.\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2.\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4.\left(1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(=182.\frac{1}{8}.\frac{808080}{919191}=\frac{182}{8}.\frac{80}{91}=20\)
19 tháng 3 2016
=\(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2.\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4.\left(1-\frac{1}{7}+\frac{1}{49}+\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\right):\frac{91}{80}\)
=\(\frac{1}{2}:4:\frac{91}{80}=\frac{10}{91}\)
bạn Bùi Minh Anh bạn không biết làm thì bạn cứ nói hẳn không ai trách bạn chứ đề này là đề đúng nhé !
NL
25 tháng 2 2020
Hình như câu này tớ đã gặp đâu đó trong đề thi HSG rồi!
\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\div\frac{4+\frac{4}{7}+\frac{4}{9}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{343}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\div\frac{4\left(1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}\right)}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}}\)
\(=\frac{1}{2}\div4=\frac{1}{8}\)
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right):\frac{919191}{808080}\)
\(=\left(\frac{1}{2}:4\right):\frac{919191}{808080}=\frac{1}{8}\cdot\frac{808080}{919191}=\frac{10}{91}\)
Bài giải
\(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)
\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)