1/4x6+1/8x9+1/12x12+1/16x15+....+1/2680x2013
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Những câu hỏi liên quan
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
28 tháng 7 2023
\(\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}\)
= \(\dfrac{1}{5}-\dfrac{1}{6}\times\dfrac{1}{6}-\dfrac{1}{7}\times\dfrac{1}{7}-\dfrac{1}{8}\times\dfrac{1}{8}-\dfrac{1}{9}\times\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{5}-\dfrac{1}{10}\)
= \(\dfrac{1}{10}\)
VN
Võ Ngọc Phương
VIP
6 tháng 9 2023
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{40.42}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{40}-\dfrac{1}{42}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{42}\right)\)
\(=\dfrac{1}{2}.\dfrac{10}{21}\)
\(=\dfrac{5}{21}\)
\(#Wendy.Dang\)
H9
HT.Phong (9A5)
CTVHS
6 tháng 9 2023
\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{40\cdot42}\)
\(=\dfrac{1}{2}\cdot\left(2\cdot\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{40\cdot42}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{40\cdot42}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{40}-\dfrac{1}{42}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{42}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{41}{42}\)
\(=\dfrac{41}{84}\)
TN
28 tháng 11 2018
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{8\times9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{1}-\frac{1}{9}=\frac{8}{9}\)
28 tháng 11 2018
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{1}-\frac{1}{9}\)
\(=\frac{8}{9}\)
B
20 tháng 3 2016
\(A=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}....\frac{1}{9x10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{9}-\frac{1}{10}=\frac{9}{10}\)
20 tháng 3 2016
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)
ML
6 tháng 7 2017
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
6 tháng 7 2017
Ta có:
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{98.100}\)
\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{98.100}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)
Vậy giá trị của biểu thức là \(\frac{49}{200}\)
19 tháng 3 2017
\(\frac{1}{2x4}\)+ \(\frac{1}{4x6}\)+ ... + \(\frac{1}{98x100}\)= \(\frac{1}{2}\)x(\(\frac{4-2}{2x4}\)+\(\frac{6-4}{4x6}\)+ ... + \(\frac{100-98}{98x100}\))
= \(\frac{1}{2}\)x(\(\frac{1}{2}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{8}\)+ ... + \(\frac{1}{98}\)-\(\frac{1}{100}\))
= \(\frac{1}{2}\)x(\(\frac{1}{2}\)-\(\frac{1}{100}\)) = \(\frac{49}{200}\)
16 tháng 3 2016
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{96.98}+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)