HH
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K
Khách
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HM
0
IM
20 tháng 8 2016
b)
\(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)
\(\frac{7}{3.4}+\frac{7}{4.5}+.....+\frac{7}{60.61}\)
\(=7\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{60}-\frac{1}{61}\right)\)
\(=7\left(\frac{1}{3}-\frac{1}{61}\right)\)
\(=\frac{406}{183}\)
d)
\(\frac{6}{2.4}+\frac{6}{4.6}+....+\frac{1}{72.74}\)
\(=3\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{72}-\frac{1}{74}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{74}\right)\)
=57/37
NH
3
22 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{7}{5}\cdot\dfrac{8}{19}+\dfrac{7}{5}\cdot\dfrac{12}{19}-\dfrac{7}{5}\cdot\dfrac{1}{18}\)
`=`\(\dfrac{7}{5}\cdot\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{18}\right)\)
`=`\(\dfrac{7}{5}\cdot\left(\dfrac{20}{19}-\dfrac{1}{18}\right)\)
`=`\(\dfrac{7}{5}\cdot\dfrac{341}{342}=\dfrac{2387}{1710}\)
6 tháng 3 2023
\(A=\dfrac{7}{1.2}+\dfrac{7}{2.3}+\dfrac{7}{3.4}+...+\dfrac{7}{2011.2012}\)
\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)\)
\(A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)\)
\(A=7\left(1-\dfrac{1}{2012}\right)=7.\dfrac{2011}{2012}=\dfrac{14077}{2012}\)
VT
2
30 tháng 11 2017
6542 tk mình nhé chúc bạn học giỏi
ai tk mình mình tk lại
PT
1
A =\(\dfrac{7}{3.4}\) + \(\dfrac{7}{4.6}\) + \(\dfrac{7}{5.8}\) + \(\dfrac{7}{6.10}\)+...+\(\dfrac{7}{60.118}\)
A = \(\dfrac{2.7}{2.3.4}\) + \(\dfrac{2.7}{2.4.6}\)+\(\dfrac{2.7}{2.5.8}\) + \(\dfrac{2.7}{2.6.10}\)+...+\(\dfrac{2.7}{2.60.118}\)
A = 7.(\(\dfrac{2}{6.4}\)+\(\dfrac{2}{8.6}\)+\(\dfrac{2}{10.8}\)+\(\dfrac{2}{12.10}\)+...+\(\dfrac{2}{120.118}\))
A = 7.(\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+\(\dfrac{2}{8.10}\)+\(\dfrac{2}{10.12}\)+...+\(\dfrac{2}{118.120}\))
A = 7.(\(\dfrac{1}{4}-\dfrac{1}{6}\)+ \(\dfrac{1}{6}-\dfrac{1}{8}\) +\(\dfrac{1}{8}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{12}\) +...+ \(\dfrac{1}{118}\) - \(\dfrac{1}{120}\))
A = 7.( \(\dfrac{1}{4}\) - \(\dfrac{1}{120}\))
A = 7.\(\dfrac{29}{120}\)
A = \(\dfrac{203}{120}\)