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`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`
\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
=\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{10.12}\)
=\(\dfrac{1}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{10.12}\right)\)
=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)
=\(\dfrac{1}{2}.\dfrac{5}{12}\)
=\(\dfrac{5}{24}\)
Dấu chấm(.)là nhân.
\(C=\dfrac{2}{5}+\dfrac{2}{45}+...+\dfrac{2}{6885}\)
\(=\dfrac{1}{2}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{85}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{84}{85}=\dfrac{42}{85}\)
(y - \(\dfrac{1}{2}\)) : \(\left(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\right)\)= \(\dfrac{1}{3}\)
(y\(-\dfrac{1}{2}\)): \(\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)= \(\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right):\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right):\dfrac{3}{10}=\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right)=\dfrac{1}{10}\)
y = \(\dfrac{3}{5}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(=\left(\frac{1}{2}+\frac{1}{20}\right)+\left(\frac{1}{6}+\frac{1}{30}\right)+\frac{1}{12}\)
\(=\frac{11}{20}+\frac{1}{5}+\frac{1}{12}\)
\(=\frac{3}{4}+\frac{1}{12}\)
\(=\frac{10}{12}=\frac{5}{6}\)
ta nhận thấy
1/2=1-1/2
1/6=1/2-1/3
1/12=1/3-1/4
1/20=1/4-1/5
1/30=1/5-1/6
1/42=1/6-1/7
ta có:
1/2+1/6+1/12+1/20+1/30+1/42
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
bn tự hiểu nha
\(\Rightarrow x+\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x=\dfrac{47}{42}-\dfrac{5}{6}=\dfrac{2}{7}\)
sau đây là phần chữa của mình:
\(=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\)
= \(\dfrac{3}{10}\)
= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{2}-\dfrac{1}{10}\)
= \(\dfrac{2}{5}\)