chứng minh :
\(\frac{4}{3\cdot5}-\frac{6}{5\cdot7}+\frac{8}{7\cdot9}-\frac{10}{9\cdot11}+.......+\frac{2016}{2015\cdot2017}>\frac{1}{6}\)
các bạn giúp nhanh mình cái muộn rồi
chứng minh :
\(\frac{4}{3\cdot5}-\frac{6}{5\cdot7}+\frac{8}{7\cdot9}-\frac{10}{9\cdot11}+.......+\frac{2016}{2015\cdot2017}>\frac{1}{6}\)
các bạn giúp nhanh mình cái muộn rồi
\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}+\frac{10}{9.11}+...+\frac{2016}{2015.2017}\)
\(=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{2017}\right)\)
\(=2.\frac{2014}{6051}\)
\(=\frac{4028}{6051}\)
\(\Rightarrow BT>\frac{1}{6}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
giúp mk nha các bn
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
Tính nhanh:
\(\frac{10}{11}:\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)
\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\frac{8}{33}=\frac{10}{11}.\frac{33}{8}\)
\(=\frac{15}{4}\)
Trả lời:
\(\frac{10}{11}\div\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{10}{11}\div\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{10}{11}\div\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{10}{11}\div\frac{8}{33}\)
\(=\frac{10}{11}\times\frac{33}{8}\)
\(=\frac{15}{4}\)
\(\frac{10}{11}:(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11})\)
=\(\frac{10}{11}:(\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+\frac{2}{9}-\frac{2}{11})\)
=\(\frac{10}{11}:(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11})\)
=\(\frac{10}{11}:(\frac{1}{3}-\frac{1}{11})\)
=\(\frac{10}{11}:\frac{8}{33}\)
= \(\frac{10}{11}.\frac{33}{8}\)= \(\frac{15}{4}\)
Tính nhanh:
\(\frac{10}{11}:\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)
\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\left(\frac{11}{33}-\frac{3}{33}\right)\)
\(=\frac{10}{11}:\frac{8}{33}\)
\(=\frac{15}{4}\)
Học tốt
\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\frac{8}{33}\)
\(=\frac{10}{11}.\frac{33}{8}\)
\(=\frac{15}{4}\)
\(\frac{10}{11}:\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\frac{8}{33}=\frac{15}{4}\)
\(\sqrt[2]{4\cdot9\frac{8}{8}+\frac{48\cdot11+5}{1\cdot\frac{814}{5+\frac{6145}{1\cdot\frac{821}{614}}}}}2548-\frac{8452}{14\cdot\frac{58}{96\cdot\frac{41}{\frac{24}{1\cdot\frac{975545}{1421+\frac{84874}{\frac{1+2+3+4+5+6+7+8+9\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot\frac{2}{1}}}}}}}}\)
A = \(\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}\)
\(A=\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}\)
\(A=\frac{2\cdot2}{3\cdot5}+\frac{2\cdot2}{5\cdot7}+\frac{2\cdot2}{7\cdot9}+\frac{2\cdot2}{9\cdot11}\)
\(A=2\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(A=2\cdot\frac{8}{33}\)
\(A=\frac{16}{33}\)
Ta có:
\(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}\)
\(A=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(A=2\cdot\frac{8}{33}\)
\(A=\frac{16}{33}\)
\(\frac{1}{10\cdot9}-\frac{1}{9\cdot8}-\frac{1}{8\cdot7}-\frac{1}{7\cdot6}-\frac{1}{6\cdot5}-\frac{1}{5\cdot4}-\frac{1}{4\cdot3}-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)
\(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
HÃY CHỨNG MINH :
\(\frac{2}{5}< A< \frac{8}{9}\)
Ta có :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\Rightarrow A< \frac{8}{9}\)(1)
Lại có \(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\Rightarrow A>\frac{2}{5}\)(2)
Từ (1) (2) => \(\frac{2}{5}< A< \frac{8}{9}\left(\text{ĐPCM}\right)\)
Bài làm :
Ta có :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>\frac{1}{2}-\frac{1}{10}\)
\(A>\frac{2}{5}\left(1\right)\)
Ta cũng có :
\( A=\frac{1}{2.2}+\frac{1}{3.3}+......+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{8.9}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-......+\frac{1}{8}-\frac{1}{9}\)
\(A< 1-\frac{1}{9}\)
\(A< \frac{8}{9}\left(2\right)\)
\(\text{Từ (1) và (2) }\Rightarrow\frac{2}{5}< A< \frac{8}{9}\)
=> Điều phải chứng minh
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(S=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
HÃY CHỨNG MINH \(\frac{2}{5}< S< \frac{7}{8}\)
Bài làm:
Ta có: \(S=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)
\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)\(\Rightarrow\frac{2}{5}< S\)
Cái còn lại tự CM