Tính 1/1.3+1/3.5+1/3.7+........+1/101.103
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HT
13 tháng 7 2015
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{101.103}\)
=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{101.103}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{101}-\frac{1}{103}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{103}\right)\)
=\(\frac{1}{2}.\frac{102}{103}\)
=\(\frac{51}{103}\)
LB
1
PN
4 tháng 11 2015
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)=\frac{1}{2}\left(1-\frac{1}{103}\right)=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)
KC
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 9
a; C = \(\dfrac{3}{1.3}\) + \(\dfrac{3}{3.5}\) + \(\dfrac{3}{3.7}\) + ... + \(\dfrac{3}{49.51}\)
C = \(\dfrac{3}{2}\).(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{49.51}\))
C = \(\dfrac{3}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{49}\) - \(\dfrac{1}{51}\))
C = \(\dfrac{3}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{51}\))
C = \(\dfrac{3}{2}\).\(\dfrac{50}{51}\)
C = \(\dfrac{25}{17}\)
J
2
NH
20 tháng 3 2023
\(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\)
\(A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2023}\\ A=\dfrac{2023}{2023}-\dfrac{1}{2023}\\ A=\dfrac{2022}{2023}\)
25 tháng 4 2019
A=1 - 1/3+1/3 - 1/5+1/5 - 1/6+...+1/99 - 101+1/101 - 1/103
A=1 - 1/103
A=102/103
KN
25 tháng 4 2019
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}\)
\(=\frac{51}{103}\)
E
5
2 tháng 4 2017
biểu thức trên = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=\frac{100}{101}< 1\)
vậy A<1
TT
3
DP
22 tháng 7 2017
\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)
\(Q=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{64}-\frac{1}{67}\)
\(Q=\frac{1}{4}-\frac{1}{67}=\frac{63}{268}\)
\(M=\frac{22}{1.3}+\frac{22}{3.5}+...+\frac{22}{101.103}\)
\(M=\frac{22}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(M=11\cdot\left(1-\frac{1}{103}\right)\)
\(M=11\cdot\frac{102}{103}=\frac{1122}{103}\)
MC
22 tháng 7 2017
\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)
\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{64}-\frac{1}{67}\right)\)
\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{67}\right)\)
\(\Leftrightarrow Q=3.\frac{63}{268}\)
\(\Leftrightarrow Q=\frac{189}{268}\)
Câu b) bạn làm tương tự nhé :)
A = \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + ... + \(\dfrac{1}{101.103}\)
A = \(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{101.103}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{101}\) - \(\dfrac{1}{103}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{103}\))
A = \(\dfrac{1}{2}\). \(\dfrac{102}{103}\)
A = \(\dfrac{51}{103}\)
Em ơi thừa số thứ ba phải là \(\dfrac{1}{5.7}\) mới đúng em nhé.