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10 tháng 9 2016
\(B=1.3+3.5+5.7+.....+95.97+97.99\)
\(\frac{2}{B}=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{95.97}+\frac{2}{97.99}\)
\(\frac{2}{B}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{97}-\frac{1}{99}\)
\(\frac{2}{B}=\frac{1}{1}-\frac{1}{99}=\frac{90}{99}=\frac{30}{33}\)
\(B=\frac{2}{\frac{30}{33}}=\frac{2.33}{30}=\frac{33}{15}\)
10 tháng 9 2016
\(6A=1.3.6+3.5.6+5.7.6+...+97.99.6\)
= \(1.3\left(5+1\right)+3.5\left(7-1\right)+5.7\left(9-3\right)+...97.99\left(101-95\right)\)
= \(.3.5+1.3+3.5.7-1.3.5+5.7.9-3.5.7+...+97.99.101-97.97.99\)
= 3 + 97 .99 . 101
= \(\frac{1+97.33.101}{2}\)
4 tháng 5 2022
\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + ..... + \(\dfrac{2}{95.97}\)
= 1 - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + .... + \(\dfrac{1}{95}\) - \(\dfrac{1}{97}\)
= \(1-\dfrac{1}{97}\)
= \(\dfrac{96}{97}\)
4 tháng 5 2022
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{95\times97}\)
\(=\dfrac{2}{3}\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{95\times97}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{97}\right)\)\(=\dfrac{2}{3}\times\dfrac{96}{97}\)\(=\dfrac{64}{97}\)
PD
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
13 tháng 10 2016
6.B=1.3.6+3.5.6+5.7.6+...+95.97.6+97.99.6
6.B=1.3.(5+1)+3.5.(7-1)+5.7.(9-3)+...+95.97.(99-93)+97.99(101-95)
6.B=1.3.5+1.3+3.5.7-1.3.5+5.7.9-3.5.7+...+95.97.99-93.95.97+97.99.101-95.97.99=1.3+97.99.101
B=(3+97.99.101)/6
LK
0
NT
2
11 tháng 4 2015
=3.(3/1.3+3/3.5+3/5.7+...+3/95.97+3/97.99)
=3(1-1/3+1/3-1/5+1/5-1/7+...+1/95-1/97+1/97-1/99)
=3[(1-1/99)+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)]
=3(1-1/99)=3(99/99-1/99)=3.98/99=1.98/33=98/33
TN
3
23 tháng 4 2019
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
\(A=\frac{1}{2}.\frac{98}{99}\)
\(A=\frac{49}{99}\)
L
23 tháng 4 2019
\(A=\frac{1}{1\cdot3} +\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{95\cdot97}+\frac{1}{97\cdot99}\)
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{95\cdot97}+\frac{2}{97\cdot99}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}\)
\(2A=\frac{98}{99}\)
\(A=\frac{98}{99}\text{ : }2\)
\(A=\frac{98}{99}\cdot\frac{1}{2}\)
\(A=\frac{49}{99}\)
DT
2
N
25 tháng 3 2017
\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\frac{100}{101}\)
\(=\frac{350}{101}\)
k mk nha
25 tháng 3 2017
7/1.3+7/3.5+7/5.7+...+7/99.101
=7(1/1.3+1/3.5+1/5.7+...+1/99.101)
=7(1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)
=7(1-1/101)
=7.100/101
=700/101
Đầy đủ ko bỏ bước nào lun!!
K CHO MK NHA!!!
BN
23 tháng 2 2017
\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
QD
23 tháng 2 2017
S=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{95.97}+\frac{1}{97.99}\)
S=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{99}\right)\)
S=\(\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
S=\(\frac{1}{2}.\frac{98}{99}\)
S=\(\frac{49}{99}\)
A=1.3+3.5+5.7+...+95.97+97.99
6A=1.3.6+3.5.6+5.7.6+...+95.97.96+97.99.96
=1.3.(5+1)+3.5.(7-1)+...+95.97.(99-93)+97.99.(101-95)
=1.1.3+1.3.5-1.3.5+3.5.7-....-95.97.99+97.99.101
=3.97.99.101
=>A=\(\frac{3+97.99.101}{6}=\frac{1+97.33.101}{2}\)\(=161651\)
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