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KT
2 tháng 8 2018
\(G=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+.....+\frac{1}{95.97.99}\)
\(=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+....+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(=\frac{1}{4}.\frac{3200}{9603}=\frac{800}{9603}\)
MT
1
17 tháng 9 2015
Nè bé
B= 1.3+ 3.5 + 5.7 +... +97.99
2:B= 2/1.3 + 2/3.5 + 2/5.7 +.... + 2/97.99
2/B= 1/1-1/3+1/3-1/5+1/5-1/7+....+1/97-1/99
2/B=1/1-1/99
2:B=98/99
B=2: 98/99
B= 2. 99/98
B= 99/49
29 tháng 9 2014
1/3*5+1/5*7+1/7*9+...+1/97*99
=1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
=1/3-1/99
32/99
KS
4
3 tháng 11 2016
a) \(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{97.99}\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{3}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{3}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{32}{99}\)
\(=\frac{16}{33}\)
3 tháng 11 2016
b)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)
\(=1-\frac{1}{103}\)
\(=\frac{102}{103}\)
20 tháng 6 2018
A= 1 x 3 + 3 x 5 + 5 x7 +...+ 95 x 97 + 97 x 99
\(\Rightarrow6A=1x3x6+3x5x6+5x7x6+...+95x97x6+97x99x6\)
\(6A=1x3x\left(5+1\right)+3x5x\left(7-1\right)+5x7x\left(9-3\right)+...+95x97x\left(99-93\right)+97x99x\left(101-95\right)\)
\(6A=1x3x5+1x3+3x5x7-1x3x5+5x7x9-3x5x7+...+95x97x99-93x95x97\)
\(+97x99x101-95x97x99\)
\(6A=\left(1x3x5+3x5x7+5x7x9+...+95x97x99+97x99x101\right)-\)
\(\left(1x3x5+3x5x7+...+93x95x97+95x97x99\right)+1x3\)
\(6A=97x99x101+1x3\)
\(6A=969903+3\)
\(6A=969906\)
\(A=\frac{969906}{6}\)
\(A=161651\)
4 tháng 8 2017
S = 1 + 3 + 5 + ... + 99
S = (1+99) + (3+97) + ... + ( 49+51)
S= 100 + 100 + ... + 100
S= 25. 100
S = 2500
A = 99 - 97 + 95 - 93 + ... + 7 - 5 + 3 - 1
A = (99-97) + (95-93) + ... + (7-5) + (3-1)
A = 2 + 2 + ... + 2 + 2
A = 2. 25
A= 50
NL
1
XO
11 tháng 8 2020
Đặt \(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right):\left(\frac{1}{1.99}+\frac{1}{3.97}+....+\frac{1}{97.3}+\frac{1}{99.1}\right)\)
Đặt \(B=\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{97.3}+\frac{1}{99.1}\)
=> 100 x B = \(\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{97.3}+\frac{100}{99.1}=1+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+...+\frac{1}{97}+\frac{1}{3}+\frac{1}{99}+1\)
=> 100 x B = \(2.\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)\)
=> \(B=\frac{1}{50}.\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)\)
Khi đó A = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{50}\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}=\frac{1}{\frac{1}{50}}=50\)
Dấu " . " là dấu nhân
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}\)
\(2A=\frac{98}{99}\)
\(A=\frac{49}{99}\)
\(B=\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{97\times99}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(B=1-\frac{1}{99}\)
\(B=\frac{98}{99}\)