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NT
2
DD
Đoàn Đức Hà
Giáo viên
7 tháng 11 2021
\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}\)
\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+\frac{2}{11\times13}\right)\)
\(=\frac{1}{2}\times\left(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+\frac{11-9}{9\times11}+\frac{13-11}{11\times13}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{13}\right)=\frac{6}{13}\)
Do đó ta có:
\(\frac{6}{13}\times y=\frac{3}{5}\)
\(\Leftrightarrow y=\frac{13}{10}\).
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 3 2023
A = \(\dfrac{4}{1\times3}\) - \(\dfrac{8}{3\times5}\) + \(\dfrac{12}{5\times7}\) - \(\dfrac{16}{7\times9}\) + \(\dfrac{20}{9\times11}\) - \(\dfrac{24}{11\times13}\)
A = ( \(\dfrac{1}{1}+\dfrac{1}{3}\)) - ( \(\dfrac{1}{3}\) + \(\dfrac{1}{5}\)) + (\(\dfrac{1}{5}\)+ \(\dfrac{1}{7}\)) - ( \(\dfrac{1}{7}\) + \(\dfrac{1}{9}\)) +( \(\dfrac{1}{9}\)+ \(\dfrac{1}{11}\)) - (\(\dfrac{1}{11}\)+\(\dfrac{1}{13}\))
A = \(\dfrac{1}{1}+\dfrac{1}{3}\) - \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}\) - \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{11}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{12}{13}\)
KV
16 tháng 3 2023
2/(7 × 9) + 2/(9 × 11) + 2/(11 × 13) + ... + 2/(97 × 99)
= 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + ... + 1/97 - 1/99
= 1/7 - 1/99
= 92/693
LJ
9 tháng 11 2017
Ta có: \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^6}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^6}}{2}\)
31 tháng 5 2018
\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}+\frac{1}{11x13}\)
\(=\frac{1}{2}x\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}x\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{1}{2}x\frac{10}{39}\)
\(=\frac{5}{39}\)
31 tháng 5 2018
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)
10 tháng 6 2017
Ta có : \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{11.13}\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+......+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}\)
\(=\frac{10}{39}\)
10 tháng 6 2017
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{11\times13}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{11\times13}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
9 tháng 6 2018
\((\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11})\cdot y=\frac{2}{3}\)
\(\Rightarrow(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11})\cdot y=\frac{2}{3}\)
\(\Rightarrow1-\frac{1}{11}\cdot y=\frac{2}{3}\)
\(\Rightarrow\frac{10}{11}\cdot y=\frac{2}{3}\)
\(\Rightarrow y=\frac{2}{3}:\frac{10}{11}=\frac{11}{15}\)
Vậy :\(y=\frac{11}{15}\)
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NT
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