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\(\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{2013.2015}=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=2.\left(\frac{2015}{2015}-\frac{1}{2015}\right)\)
\(=2.\frac{2014}{2015}\)
\(=\frac{4028}{2015}\)
4/3.5+4/5.7+4/7.9+4/9.11
=4.(1/3.5+1/5.7+1/7.9+1/9.11)
=4.1/2.(2/3.5+2/5.7+2/7.9+2/9.11)
=2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)
=2.(1/3-1/11)
=2.8/33
=16/33
4/3.5+4/5.7+4/7.9+4/9.11
=4.2/2.3.5+4.2/2.5.7+4.2/2.7.9+4.2/2.9.11
=4/2.2/3.5+4/2.2/5.7+4/2.2/7.9+4/2.2/9.11
=4/2.(2/3.5+2/5.7+2/7.9+2/9.11)
=4/2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)
=2.(1/3-1/11)
=2.8/33
=16/33
\(2B=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{23}-\frac{1}{25}\right)\)
\(2B=2\left(\frac{1}{3}-\frac{1}{25}\right)\)
\(2B=2\times\frac{22}{75}\)
\(B=\frac{44}{75}\)
4/5.7+4/7.9+...+4/59.61
=2.(2/5.7+2/7.9+...+2/59.61)
=2.(1/5-1/7+1/7-1/9+...+1/59-1/61)
=2.(1/5-1/61)
=2.56/305
=112/203
A=( 2/5.7+2/7.9+.........+2/59.61).2
A = (1/5-1/7+1/7-1/9+.......+1/59-1/61).2
A= ( 1/5-1/61)2
\(=4\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{55}\right)\)
\(=4.\frac{2}{11}\)
\(=\frac{8}{11}\)
\(\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+...+\frac{4}{97\cdot99}\)
\(=\frac{2\cdot2}{3\cdot5}+\frac{2\cdot2}{5\cdot7}+\frac{2\cdot2}{7\cdot9}+...+\frac{2\cdot2}{97\cdot99}\)
\(=\frac{2}{3}+\frac{2}{5}-\frac{2}{5}+\frac{2}{7}-\frac{2}{7}+\frac{2}{9}-...+\frac{2}{97}-\frac{2}{99}\)
\(=\frac{2}{3}-\frac{2}{99}\)
\(=\frac{64}{99}\)
Ta có :
4/5 . 7 + 4/7 . 9 + ...+ 4/59 . 61
= 2 . ( 2/5 . 7 + 2/7 . 9 + ...+ 2/59 . 61 )
= 2 . ( 1/5 - 1/7 + 1/7 - 1/9 + ...+ 1/59 - 1/61 )
= 2 . ( 1/5 - 1/61 )
= 2 . 56/305
= 112/305
Tham khảo nha !!!
`A=4/[5.7]+4/[7.9]+4/[9.11]+...+4/[21.23]+4/[23.25]`
`A=2.(2/[5.7]+2/[7.9]+....+2/[23.25])`
`A=2.(1/5-1/7+1/7-1/9+....+1/23-1/25)`
`A=2.(1/5-1/25)`
`A=2. 4/25`
`A=8/25`