Question

\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)

Answer 1

`1/8+1/24+1/48+1/80+1/120`

`=1/[2xx4]+1/[4xx6]+1/[6xx8]+1/[8xx10]+1/[10xx12]`

`=1/2xx(2/[2xx4]+2/[4xx6]+2/[6xx8]+2/[8xx10]+2/[10xx12])`

`=1/2xx(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)`

`=1/2xx(1/2-1/12)`

`=1/2xx(6/12-1/12)`

`=1/2xx5/12=5/24`

Answer 2

\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)

=\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{10.12}\)

=\(\dfrac{1}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{10.12}\right)\)

=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)

=\(\dfrac{1}{2}.\dfrac{5}{12}\)

=\(\dfrac{5}{24}\)

Dấu chấm(.)là nhân.

Answer 3

   \(\dfrac{1}{8}\) + \(\dfrac{1}{24}\)+ \(\dfrac{1}{48}\)+ \(\dfrac{1}{80}\)+ \(\dfrac{1}{120}\)

 = \(\dfrac{1}{2}\) X (\(\dfrac{2}{2\times4}\)+ \(\dfrac{2}{4\times6}\)+\(\dfrac{2}{6\times8}\)+\(\dfrac{2}{8\times10}\)+ \(\dfrac{2}{10\times12}\))

= \(\dfrac{1}{2}\) x ( \(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{12}\))

= \(\dfrac{1}{2}\)x (\(\dfrac{1}{2}\)-\(\dfrac{1}{12}\))

= \(\dfrac{1}{2}\) X \(\dfrac{6-1}{12}\)

= \(\dfrac{1}{2}\)x \(\dfrac{5}{12}\)

= \(\dfrac{5}{24}\)

Answer 4

$\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}$

 

$=\frac{1}{2\times 4}+\frac{1}{4\times 6}+\frac{1}{6\times 8}+\frac{1}{8\times 10}+\frac{1}{10\times 12}$

 

$=\frac{1}{2}\times (\frac{2}{2\times 4}+\frac{2}{4\times 6}+\frac{2}{6\times 8}+\frac{2}{8\times 10}+\frac{2}{10\times 12})$

 

$=\frac{1}{2}\times (\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12})$

 

$=\frac{1}{2}\times (\frac{1}{2}-\frac{1}{12})$

 

$=\frac{1}{2}\times (\frac{6}{12}-\frac{1}{12})$

 

$=\frac{1}{2}\times \frac{5}{12}$

 

$=\frac{5}{24}$