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TH
0
Q
1
19 tháng 8 2018
\(2\cdot\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right)\)
Theo quy luật :\(2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(2.\left(1-\frac{1}{15}\right)\)
\(2.\frac{14}{15}\)
\(\frac{28}{15}\)
MT
0
NT
0
DB
3
VN
Võ Ngọc Phương
VIP
8 tháng 8 2023
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
T
22 tháng 10 2023
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{13\times15}+\dfrac{2}{15\times17}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}\)
\(=1-\dfrac{1}{17}\)
\(=\dfrac{16}{17}\)
22 tháng 10 2023
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{15\cdot17}\)
\(=2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{17}\)
\(=2-\dfrac{1}{17}\)
\(=\dfrac{33}{17}\)
HT
27 tháng 10 2020
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
KN
3 tháng 1 2019
\(\text{Đặt }A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}+\frac{1}{15.17}\)
\(\Leftrightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{15.17}\)
\(\Leftrightarrow2A=\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+\frac{17-15}{15.17}\)
\(\Leftrightarrow2A=\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+...+\frac{17}{15.17}-\frac{15}{15.17}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}+...+\frac{1}{15}-\frac{1}{17}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{7}\)
\(\Leftrightarrow2A=\frac{4}{21}\)
\(\Rightarrow A=\frac{4}{21}\times\frac{1}{2}\)
\(\Rightarrow A=\frac{2}{21}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 8 2023
Bài 1:
A = \(\dfrac{1}{1\times3}\) + \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) +...+ \(\dfrac{1}{2019\times2021}\)
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{1\times3}\) + \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+...+ \(\dfrac{2}{2019\times2021}\))
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+...+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2021}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2021}\))
A = \(\dfrac{1010}{2021}\)
\(2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{13.15}\right)=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{13.15}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}=1-\frac{1}{15}=\frac{14}{15}\)
\(=2.2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=4.\left(1-\frac{1}{15}\right)\)
\(=4.\frac{14}{15}\)
\(=\frac{56}{15}\)