\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\) giúp mình với
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7 tháng 8 2016
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
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 ^^!!
5 tháng 3 2016
\(-\frac{2}{1.3}-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-\frac{2}{9.11}-\frac{2}{11.13}-\frac{2}{13.15}\)
\(=\left(-\frac{2}{1.3}\right)+\left(-\frac{2}{3.5}\right)+\left(-\frac{2}{5.7}\right)+\left(-\frac{2}{7.9}\right)+\left(-\frac{2}{9.11}\right)+\left(-\frac{2}{11.13}\right)+\left(-\frac{2}{13.15}\right)\)
\(=\left(-2\right).\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\right)\)
\(=\left(-2\right).\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\left(-2\right).\left(1-\frac{1}{15}\right)=\left(-2\right).\frac{14}{15}\)
\(=-\frac{28}{15}\)
21 tháng 11 2015
M=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
M=\(\frac{1}{3}-\frac{1}{99}\)
M=\(\frac{32}{99}\)
TICK ỦNG HỘ NHA
10 tháng 7 2019
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
10 tháng 7 2019
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
3 tháng 4 2018
1/5.6 + 1/6.7 + 1/7.8 +...+ 1/24.25
=1/5 - 1/6 + 1/6-1/7 +1/7-1/8 + ... + 1/24-1/25
=> Kết quả là: 1/5 - 1/25 = 4/25
b) 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+...+ 2/99.101
=2/1-2/3 + 2/3-2/5 + 2/5-2/7 + 2/7-2/9 + ... + 2/99-2/101
=> kết quả là 2/1 - 2/101 =200/101
3 tháng 4 2018
a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
=\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
=\(\frac{1}{5}-\frac{1}{25}\)
=\(\frac{4}{25}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
=\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
=\(2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
=\(2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
=\(2.\frac{100}{101}\)
=\(\frac{200}{101}\)
16 tháng 7 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\frac{8}{9}\)
\(=\frac{4}{9}\)
16 tháng 7 2016
Đặt: A=1/1.3+1/3.5+1/5.7+1/7.9
2A=2/1.3+2/3.5+2/5.7+2/7.9
2A=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9
2A=1-1/9
2A=8/9
A=4/9
4 tháng 2 2017
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\)
\(=1-\left(\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}\right)\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
2/1.3+2/3.5+2/5.7+2/7.9
=1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9
=1/1-1/9
=8/9