BC
5
K
Khách
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5 tháng 3 2016
\(-\frac{2}{1.3}-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-\frac{2}{9.11}-\frac{2}{11.13}-\frac{2}{13.15}\)
\(=\left(-\frac{2}{1.3}\right)+\left(-\frac{2}{3.5}\right)+\left(-\frac{2}{5.7}\right)+\left(-\frac{2}{7.9}\right)+\left(-\frac{2}{9.11}\right)+\left(-\frac{2}{11.13}\right)+\left(-\frac{2}{13.15}\right)\)
\(=\left(-2\right).\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\right)\)
\(=\left(-2\right).\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\left(-2\right).\left(1-\frac{1}{15}\right)=\left(-2\right).\frac{14}{15}\)
\(=-\frac{28}{15}\)
1 tháng 2 2017
\(\frac{-2}{1.3}-\frac{2}{3.5}-...-\frac{2}{13.15}\)
\(=-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{13.15}\right)\)
\(=-\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=-\left(1-\frac{1}{15}\right)\)
\(=\frac{-14}{15}\)
4 tháng 4 2017
B=2(2/3.5 - 2/ 5.7 +....................+ 2/99.101)
B=2(1/3.5 -2/5.7+..............+1/99.100)
B=2(1/3-1/5+1/5-.............+1/99-1/100)
B=2(1/3-1/100)
B=2.97/100
B=97/50
2 tháng 7 2015
1)
2/3.5+2/5.7+...+2/11.13+2/13.15+2/1.2+2/2.3+...+2/9.10
=(2/3.5+...2/13.15)+(2/1.2+...+2/9.10)
= (2/3-2/15)+ [2(1-1/10)]
=8/15+9/5
=7/3
2)
11/12+11/12.24+...+11/88.99
=11-1/9
=10/8/9
LL
3
PA
30 tháng 7 2016
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
30 tháng 7 2016
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
NQ
0
26 tháng 8 2018
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) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{35.37}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{35}-\frac{1}{37}\)
\(=\frac{1}{3}-\frac{1}{37}=\frac{34}{111}\)
26 tháng 8 2018
c) \(\frac{7}{7.9}+\frac{7}{9.11}+\frac{7}{11.13}+...+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{101}\right)=\frac{7}{2}\cdot\frac{94}{707}=\frac{47}{101}\)
25 tháng 2 2018
\(B=\frac{2^3}{3.5}+\frac{2^3}{5.7}+....+\frac{2^3}{101.103}\)
\(\Rightarrow\frac{1}{2^2}.B=\frac{2}{3.5}+\frac{2}{4.7}+....+\frac{2}{101.103}\)
\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\)
\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow B=\frac{100}{309}:\frac{1}{4}=\frac{400}{309}\)
S
25 tháng 2 2018
\(=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(=4\cdot\frac{100}{309}=\frac{400}{309}\)
hình như đề bài bị thiếu rồi bn ơi
\(\frac23.5+\frac25.7+\frac27.11+\frac{2}{11}.3\)
= \(\frac{10}{3}+\frac{14}{5}+\frac{22}{7}+\frac{6}{11}\)
= \(\frac{92}{15}\) + \(\frac{344}{187}\)
= \(7,9729.\ldots\)