Cho em hỏi: 6/3.5 + 6/5.7+.......+6/2009.2011
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H
1
3 tháng 3 2022
\(\dfrac{6}{1.3}+\dfrac{6}{3.5}+...+\dfrac{6}{99.100}\\ =3\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.100}\right)\\ =3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(1-\dfrac{1}{100}\right)\\ =3.\dfrac{99}{100}\\ =\dfrac{297}{100}\)
LN
1
HT
11 tháng 8 2015
C = \(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
C = \(2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
C = \(2.\left(\frac{1}{15}-\frac{1}{90}\right)=2.\frac{1}{18}\)
C = \(\frac{1}{9}\)
MT
11 tháng 8 2015
\(B=\frac{6}{1.3}+\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{99.101}\)
\(=3.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{9}{99.101}\right)\)
\(=3.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=3.\left(\frac{1}{1}-\frac{1}{101}\right)=3.\left(\frac{101}{101}-\frac{1}{101}\right)=3.\frac{100}{101}=\frac{300}{101}\)
\(C=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)
\(=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+....+\frac{1}{87}-\frac{1}{90}\right)\)
\(=2.\left(\frac{1}{15}-\frac{1}{90}\right)=2.\left(\frac{6}{90}-\frac{1}{90}\right)=2.\frac{5}{90}=\frac{1}{9}\)
MV
10 tháng 2 2018
\(\dfrac{6}{1\cdot3}+\dfrac{6}{3\cdot5}+...+\dfrac{6}{\left(n-2\right)n}\\ =3\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(n-2\right)n}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}\right)\\ =3\left(1-\dfrac{1}{n}\right)\\ =3\cdot\dfrac{n-1}{n}\)
DB
0
HT
2
9 tháng 5 2016
B=\(\frac{6}{1.3}+\frac{6}{3.5}+\frac{6}{5.7}+......+\frac{6}{99.101}\)
=\(6.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.......+\frac{1}{99.101}\right)\)
=\(6\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{99}-\frac{1}{101}\right)\)
=\(6.\left(1-\frac{1}{101}\right)\)
=\(6.\frac{100}{101}\)
=\(\frac{600}{101}\)
NT
0
\(\frac{6}{3\cdot5}+\frac{6}{5\cdot7}+....+\frac{6}{2009\cdot2011}\)
\(=3\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2009\cdot2011}\right)\)
\(=3\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=3\cdot\left(\frac{1}{3}-\frac{1}{2011}\right)\)
\(=3\cdot\left(\frac{2011}{6033}-\frac{3}{6033}\right)\)
\(=3\cdot\frac{2008}{6033}\)
\(=\frac{2008}{2011}\)