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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
\(B=\frac{6}{1.3}+\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{99.101}\)
\(=3.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{9}{99.101}\right)\)
\(=3.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=3.\left(\frac{1}{1}-\frac{1}{101}\right)=3.\left(\frac{101}{101}-\frac{1}{101}\right)=3.\frac{100}{101}=\frac{300}{101}\)
\(C=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)
\(=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+....+\frac{1}{87}-\frac{1}{90}\right)\)
\(=2.\left(\frac{1}{15}-\frac{1}{90}\right)=2.\left(\frac{6}{90}-\frac{1}{90}\right)=2.\frac{5}{90}=\frac{1}{9}\)
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
12 tháng 3 2023
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2021.2023}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{2021.2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}.\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
12 tháng 3 2023
Ta có A = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2021\cdot2023}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}+\dfrac{1}{2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
25 tháng 4 2017
=3(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+..+1/49-1/51)
=3x50/51=50/17
theo tôi là thế còn các bn
25 tháng 4 2017
M = \(3\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{49\cdot51}\right)\)\(=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
M = \(3\left(1-\frac{1}{51}\right)=3\cdot\frac{50}{51}=\frac{50}{17}\)
AN
3
11 tháng 4 2017
hình như là sai đề nếu đúng thì phải: \(\dfrac{6}{1.3}+\dfrac{6}{3.5}+\dfrac{6}{5.7}+\dfrac{6}{7.9}\)
S
11 tháng 4 2017
Sửa đề:
\(\dfrac{6}{1.3}+\dfrac{6}{3.5}+\dfrac{6}{5.7}+\dfrac{6}{7.9}\)
Lời giải:
Đặt \(A=\dfrac{6}{1.3}+\dfrac{6}{3.5}+\dfrac{6}{5.7}+\dfrac{6}{7.9}\)
\(\Rightarrow A=3.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}\right)\)
\(\Rightarrow A=3.\left(2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)\)
\(\Rightarrow A=3.\left(2-\dfrac{1}{9}\right)\)
\(\Rightarrow A=3.\dfrac{17}{9}\)
\(\Rightarrow A=\dfrac{17}{3}\)
\(\frac{6}{3\cdot5}+\frac{6}{5\cdot7}+....+\frac{6}{2009\cdot2011}\)
\(=3\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2009\cdot2011}\right)\)
\(=3\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=3\cdot\left(\frac{1}{3}-\frac{1}{2011}\right)\)
\(=3\cdot\left(\frac{2011}{6033}-\frac{3}{6033}\right)\)
\(=3\cdot\frac{2008}{6033}\)
\(=\frac{2008}{2011}\)