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
19 tháng 6 2018
Q=\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{89}{90}\)
Q=\(Q=\frac{5}{2.3}+\frac{11}{3.4}+\frac{19}{4.5}+\frac{29}{5.6}+\frac{41}{6.7}+\frac{55}{7.8}+\frac{71}{8.9}+\frac{89}{9.10}\)
Q=
4 tháng 8 2016
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9-\left(1-\frac{1}{10}\right)=9-1+\frac{1}{10}=8\frac{1}{10}\)
NK
23 tháng 2 2016
\(\frac{1}{2}+\frac{5}{6}+...+\frac{55}{56}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{56}\right)\)
\(=7.1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}\right)\)
\(=7-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\right)\)
\(=7-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\right)\)
\(=7-\left(1-\frac{1}{8}\right)\)
\(=7-\frac{7}{8}=\frac{49}{8}\)
6 tháng 3 2016
A=\(\frac{1}{2}\)+\(\frac{5}{6}\)+\(\frac{11}{12}\)+\(\frac{19}{20}\)+\(\frac{29}{30}\)+\(\frac{41}{42}\)=(1-\(\frac{1}{2}\))+(1-\(\frac{1}{6}\))+(1-\(\frac{1}{12}\))+(1-\(\frac{1}{20}\))+(1-\(\frac{1}{30}\))+(1-\(\frac{1}{42}\))
=1-1+\(\frac{1}{2}\)+1-\(\frac{1}{2}\)+\(\frac{1}{3}\)+1-\(\frac{1}{3}\)+\(\frac{1}{4}\)+1-\(\frac{1}{4}\)+\(\frac{1}{5}\)+1-\(\frac{1}{5}\)+\(\frac{1}{6}\)+1-\(\frac{1}{6}\)+\(\frac{1}{7}\)
=(1-1+1+1+1+1+1)+(\(\frac{1}{2}\)-\(\frac{1}{2}\))+(\(\frac{1}{3}\)-\(\frac{1}{3}\))+(\(\frac{1}{4}\)-\(\frac{1}{4}\))+(\(\frac{1}{5}\)-\(\frac{1}{5}\))+(\(\frac{1}{6}\)-\(\frac{1}{6}\))+\(\frac{1}{7}\)
=5+\(\frac{1}{7}\)=\(\frac{36}{7}\)
LC
5 tháng 8 2015
\(A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
=>\(A=\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
=>\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
=>\(A=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
=>\(a=9-\left(1-\frac{1}{10}\right)=\frac{90}{10}-\frac{9}{10}=\frac{81}{10}\)
5 tháng 8 2015
9 - A = \(1-\frac{1}{2}+1-\frac{5}{6}+1-\frac{11}{12}+..+1-\frac{89}{90}=\frac{1}{2}+\frac{1}{6}+..+\frac{1}{90}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)
=> A = \(9-\frac{9}{10}=\frac{81}{10}\)
\(x+\frac{11}{30}=\frac{19}{20}\)
\(x=\frac{19}{20}-\frac{11}{30}\)
\(x=\frac{57}{60}-\frac{22}{60}\)
\(x=\frac{35}{60}=\frac{7}{12}\)
\(x+\frac{11}{30}=\frac{19}{20}\)
\(x=\frac{19}{20}-\frac{11}{30}\)
\(x=\frac{7}{12}\)