Tính nhanh: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{77.79}\)
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NT
2
7 tháng 8 2016
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
QD
2 tháng 6 2016
a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
dấu / là phần nhé. bạn có thể xem bài có dấu phần ở : Câu hỏi của Nguyễn Thị Hoài Anh
2 tháng 6 2016
A)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
=1-\(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=1-\(\frac{1}{101}\)
=\(\frac{100}{101}\)
B) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{1}{99.101}\)
=5.(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{2}{2}.\)(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{1}{2}\).(\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{1}{2}\).(1-\(\frac{1}{3}\)+\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=5.\(\frac{1}{2}\).(1-\(\frac{1}{101}\))
=\(\frac{5}{2}.\frac{100}{101}=\frac{250}{100}\)
Chúc bạn học tốt
7 tháng 5 2016
\(a,=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(b,=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
7 tháng 5 2016
a,\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)=1.\frac{99}{100}=\frac{99}{100}\)
17 tháng 4 2016
a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
15 tháng 4 2017
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
LT
14 tháng 8 2017
a) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
= \(\frac{2}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
= 1. \(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
= 1. \(\left(1-\frac{1}{101}\right)\)
= 1. \(\left(\frac{101}{101}-\frac{1}{101}\right)\)
= 1. \(\frac{100}{101}\)
= \(\frac{100}{101}\)
b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
= \(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
= \(\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
= \(\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
= \(\frac{5}{2}.\left(\frac{101}{101}-\frac{1}{101}\right)\)
= \(\frac{5}{2}.\frac{100}{101}\)
= \(\frac{500}{202}\)
MD
6
VN
30 tháng 4 2016
\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{99x101}\)
\(=1-\frac{1}{3}+\frac{1}{3}+\frac{1}{5}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}+\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
30 tháng 4 2016
nếu ko làm đc thì mình làm cho nha
27 tháng 4 2015
a) \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\left(1-\frac{1}{101}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)=\left(\frac{101}{101}-\frac{1}{101}\right)+0+...+0=\frac{100}{101}\)
b) \(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}=2\cdot\frac{1}{2}\left(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{99\cdot101}\right)\)
\(=5\cdot\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\right)=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)\(=\frac{5}{2}\left[\left(1-\frac{1}{101}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)\right]\)
\(=\frac{5}{2}\left[\left(\frac{101}{101}-\frac{1}{101}\right)+0+...+0\right]=\frac{5}{2}\cdot\frac{100}{101}=\frac{5\cdot100}{2\cdot101}=\frac{5\cdot50}{1\cdot101}=\frac{250}{101}\)
Mình ko chắc là đúng đâu, do nhẩm
chúc bạn học tốt!^_^
MC
4 tháng 5 2016
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
26 tháng 4 2017
Ta có:
A = \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ \(\frac{2}{7.9}\) + ... + \(\frac{2}{2001.2003}\) + \(\frac{2}{2003.2005}\)
= \(\frac{1}{1}\) - \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{9}\) + ... + \(\frac{1}{2001}\)- \(\frac{1}{2003}\)+ \(\frac{1}{2003}\)- \(\frac{1}{2005}\)
= 1 - \(\frac{1}{2005}\)
= \(\frac{2004}{2005}\)
Chúc bạn học tốt nha ^^!!
(2/1+2/3) + (2/3+2/5) + (2/5+2/7) + ...+ (2/77+2/79) 2/1 - 2/79 156/79