NT
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
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
\(=182.\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2.\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4.\left(1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(=182.\frac{1}{8}.\frac{808080}{919191}=\frac{182}{8}.\frac{80}{91}=20\)
H
8 tháng 3 2018
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right):\frac{919191}{808080}\)
\(=\left(\frac{1}{2}:4\right):\frac{919191}{808080}=\frac{1}{8}\cdot\frac{808080}{919191}=\frac{10}{91}\)
F
29 tháng 6 2019
Bài giải
\(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)
\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)
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
\(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)
\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)
NL
25 tháng 2 2020
Hình như câu này tớ đã gặp đâu đó trong đề thi HSG rồi!
\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\div\frac{4+\frac{4}{7}+\frac{4}{9}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{343}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\div\frac{4\left(1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}\right)}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}}\)
\(=\frac{1}{2}\div4=\frac{1}{8}\)
2 tháng 5 2018
\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)
\(=\frac{2}{4}=\frac{1}{2}\)
\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)
\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)
\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)
\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)
\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\) \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\) \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\) \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)
\(=\frac{58}{7}-\frac{487}{63}\) \(=\frac{577}{45}-\frac{280}{45}\)
\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\) \(=\frac{33}{5}\)
\(P=M-N\)
\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)
\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)
\(\Rightarrow P=\frac{-272}{45}\)
Vậy P = \(\frac{-272}{45}\)
\(6)E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(=\frac{5}{11}+\frac{5}{22}-\left(10101.\frac{4}{111111}\right)\)
\(=\frac{10}{22}+\frac{5}{22}-\frac{4}{11}\)
\(=\frac{15}{22}-\frac{8}{22}=\frac{7}{22}\)
\(7)F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}.\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{64}\right)}{1\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{16}{64}-\frac{4}{64}+\frac{1}{64}-\frac{1}{256}\right)}{1\left(\frac{64}{64}-\frac{16}{64}+\frac{4}{64}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{13}{64}-\frac{1}{256}\right)}{1.\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{52}{256}-\frac{1}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{51}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{\frac{153}{256}}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{153}{256}:\frac{51}{64}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3}{4}+\frac{5}{8}\)
\(=\frac{3}{8}+\frac{5}{8}=1\)
Xin lỗi tớ đã làm hết buổi tối mà chỉ có 7 bài mong bạn thông cảm cho mình nhé !
28 tháng 3 2018
\(\left[9-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
có 9 số 1 có 9 số hạng
\(=\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=1\)
Gọi biểu thức đó là A
Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
=>\(2A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\)
\(A=1-\frac{1}{2^9}\)
A= 511/512