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a ) 1/2 .4 + 1/4 . 6 + 1/6 . 8 + .........+ 1/98 . 100
= 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ........+ 1/98 - 1/100
= 1/2 - 1/100
= 49/100
b ) 1/1 . 5 + 1/5 . 9 + 1/9 . 13 + ......+ 1/201 . 205
= 1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13+ ..... + 1/201 - 1/205
= 1 - 1/205
= 204/205
c ) 6/3 . 5 + 6/5 . 7 + 6/7 . 9 + ...... + 6/97 . 99
= 6/3 - 6/5 + 6/5 - 6/7 + 6/7 -6/9 + ........ + 6/97 - 6/99
= 6/3 - 6/99
= 64/33
d ) 4/8 . 11 + 4/11 . 14 + 4/14 . 17 + ......... 4/98 . 101
= 4/8 - 4/11 + 4/11 - 4/14 + 4/14 - 4/17 + .......+ 4/98 - 4/101
= 4/8 - 4/101
= 93/202
a) \(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+....+\frac{2}{98\times100}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
= \(\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\frac{98}{200}=\frac{49}{200}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)
- A ở trên giữa các phân số là dấu " + " nha mấy bạn !
Ta có:
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+\frac{1}{5.7}+\frac{1}{6.8}+\frac{1}{7.9}+\frac{1}{8.10}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}....+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}=\frac{1}{2}.\left(\frac{8}{9}+\frac{2}{5}\right)=\frac{1}{2}.\frac{58}{45}=\frac{29}{45}\)
\(A=\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+.........+\frac{1}{96\cdot98}\right)-\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.......+\frac{1}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+....+\frac{2}{96\cdot98}\right)-\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{96}-\frac{1}{98}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{98}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{12}{49}-\frac{16}{99}=\frac{404}{4851}\)
a)\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\cdot\frac{49}{100}\)
\(=\frac{49}{200}\)
b)\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{201}-\frac{1}{205}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{205}\right)\)
\(=\frac{1}{4}\cdot\frac{204}{205}\)
\(=\frac{51}{205}\)
c)\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=3\cdot\frac{32}{99}\)
\(=\frac{32}{33}\)
d)tương tự bạn nhân với 4/3 nhé