Chứng minh rằng: \(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}< \frac{3}{4}\)
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HT
8 tháng 8 2015
\(\frac{1}{3.1}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
= \(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}.\frac{98}{99}-\frac{1}{2}.\frac{49}{100}\)
= \(\frac{49}{99}-\frac{49}{200}\)
= \(\frac{4949}{19800}\)
8 tháng 8 2015
bn zô xem nha, ko hiểu thì cứ hỏi bn ấy nhá
http://olm.vn/hoi-dap/question/154321.html
YY
13 tháng 7 2018
\(A=\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+.........+\frac{1}{96\cdot98}\right)-\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.......+\frac{1}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+....+\frac{2}{96\cdot98}\right)-\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{96}-\frac{1}{98}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{98}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{12}{49}-\frac{16}{99}=\frac{404}{4851}\)
NT
0
TM
17 tháng 7 2016
=>\(T=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{98^2}{97.99}.\frac{99^2}{98.100}\)
=>\(T=\frac{2^2.3^2.4^2...98^2.99^2}{1.3.2.4.3.5...97.99.98.100}\)
Trông thì khó vậy nhưng thực ra ko khó đâu, bạn chỉ việc rút gọn từ trên tử xuống dưới mẫu là xong
=>\(T=\frac{2.99}{1.100}=\frac{99}{50}=1\frac{49}{50}\)
GN
17 tháng 7 2016
\(=\frac{2.2}{1.3}.\frac{3.3}{3.5}....\frac{98.98}{97.99}.\frac{99.99}{98.100}\)
\(=\frac{2.3.4....98.99}{1.3.5...97.98}.\frac{2.3.4....98.99}{3.5.7...99.100}\)
rút gọn đi có :
\(\frac{99}{1}.\frac{2}{100}=99.\frac{1}{50}=\frac{99}{50}\)
TL
1
S
20 tháng 6 2018
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}-\frac{1}{2}\cdot\frac{49}{100}\)
\(=\frac{1}{2}\left(\frac{98}{99}-\frac{49}{100}\right)=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
PA
3 tháng 12 2016
\(A=\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\frac{1}{4\times6}+\frac{1}{5\times7}+\frac{1}{6\times8}+\frac{1}{7\times9}+\frac{1}{8\times10}\)
\(2A=\frac{2}{1\times3}+\frac{2}{2\times4}+\frac{2}{3\times5}+\frac{2}{4\times6}+\frac{2}{5\times7}+\frac{2}{6\times8}+\frac{2}{7\times9}+\frac{2}{8\times10}\)
\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+\frac{1}{6}-\frac{1}{8}+\frac{1}{7}-\frac{1}{9}+\frac{1}{8}-\frac{1}{10}\)
\(2A=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\)
\(2A=\frac{58}{45}\)
\(A=\frac{58}{45}\div2\)
\(A=\frac{29}{45}\)
LT
3 tháng 12 2016
\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{10}\)
\(=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}=\frac{58}{45}\)
\(A=\frac{29}{45}\)
8 tháng 9 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)