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PT
2
MT
19 tháng 6 2015
2/15+2/35+2/63+.....+2/9603=2/3.5+2/5.7+2/7.9+...2/97.99
=1/3-1/5+1/5-1/7+...+1/97-1/99
=1/3-1/99
=33/99-1/99
=32/99
19 tháng 6 2015
\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{9603}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
DP
12 tháng 7 2017
\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}\)
\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+...+\left(1-\frac{1}{9999}\right)\)
\(=\left(1-\frac{1}{1\cdot3}\right)+\left(1-\frac{1}{3\cdot5}\right)+\left(1-\frac{1}{5\cdot7}\right)+\left(1-\frac{1}{7\cdot9}\right)+...+\left(1-\frac{1}{99\cdot101}\right)\)
\(=\left(1+1+1+1+...+1\right)-\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
Có tất cả : (101 - 3) : 2 + 1 = 50 chữ số 1 => (1 + 1 + 1 + .... + 1) = 1 x 50 = 50
\(\Rightarrow50-\frac{1}{2}\cdot\left(1-\frac{1}{101}\right)\)
\(=50-\frac{1}{2}\cdot\frac{100}{101}=50-\frac{100}{101}=\frac{4950}{101}\)
Vậy \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}=\frac{4950}{101}\)
AH
15 tháng 8 2016
- \(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\)
\(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)
\(4.B=1-\frac{1}{97}\)
\(4.B=\frac{96}{97}\)
\(B=\frac{96}{97}:4\)
\(B=\frac{24}{97}\)
5 tháng 1 2016
\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+...+\frac{2}{899}\)
\(=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{29\cdot31}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{31}\)
\(=\frac{1}{3}-\frac{1}{31}\)
\(=\frac{28}{93}\)
26 tháng 6 2017
\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{899}\)
= \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{29.31}\)
= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{31}\)
= \(\frac{1}{3}-\frac{1}{31}+0+0+...+0\)
= \(\frac{29}{93}\)
AP
2
AM
1 tháng 7 2015
\(\frac{2^2}{15}+\frac{2^2}{35}+\frac{2^2}{63}+\frac{2^2}{99}+\frac{2^2}{143}=2\cdot\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\right)=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)=2\cdot\left(\frac{1}{3}-\frac{1}{13}\right)=2\cdot\frac{10}{39}=\frac{20}{39}\)
1 tháng 7 2015
\(=2\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=2\left(1-\frac{1}{13}\right)=2.\frac{12}{13}=\frac{24}{13}\)
4 tháng 9 2015
A = 1/2 + 1/3 + 1/6 + 1/12 + 1/15 + 1/20 + 1/30 + 1/35 + 1/42 + 1/56 + 1/63 + 1/72 + 1/99
= ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 ) + ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
= ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9 ) + ( 1/1.3 + 1/3.5 + ... + 1/9.11 )
= ( 1 - 1/2 + 1/2 - 1/3 + ... + 1/8 - 1/9 ) + 1/2 . ( 2/1.3 + 2/3.5 + ... + 2/9.11 )
= ( 1 - 1/9 ) + 1/2 . ( 1 - 1/3 + 1/3 - 1.5 + ... + 1/9 - 1/11 )
= 8/9 + 1/2 . ( 1 - 1/11 )
= 8/9 + 1/2 . 10/11
= 8/9 + 5/11
= 133/99
4 tháng 9 2015
A = 1/2 + 1/3 + 1/6 + 1/12 + 1/15 + 1/20 + 1/30 + 1/35 + 1/42 + 1/56 + 1/63 + 1/72 + 1/99
= ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 ) + ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
= ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9 ) + ( 1/1.3 + 1/3.5 + ... + 1/9.11 )
= ( 1 - 1/2 + 1/2 - 1/3 + ... + 1/8 - 1/9 ) + 1/2 . ( 2/1.3 + 2/3.5 + ... + 2/9.11 )
= ( 1 - 1/9 ) + 1/2 . ( 1 - 1/3 + 1/3 - 1.5 + ... + 1/9 - 1/11 )
= 8/9 + 1/2 . ( 1 - 1/11 )
= 8/9 + 1/2 . 10/11
= 8/9 + 5/11
= 133/99
3 tháng 5 2015
\(M=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+...+1-\frac{1}{9999}\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)
\(M=50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(M=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(M=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{5050-100}{101}=\frac{4950}{101}\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
115/126