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=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
\(A=\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{223}+\frac{4}{437}+\frac{2}{575}\)
\(=\frac{2}{5.7}+\frac{2}{7.11}+\frac{2}{11.13}+\frac{4}{13.17}+\frac{2}{17.19}+\frac{4}{19.23}+\frac{2}{23.25}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
\(\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{323}+\frac{4}{437}+\frac{2}{575}\)
\(=\frac{2}{5.7}+\frac{4}{7.11}+\frac{2}{11.13}+\frac{4}{13.17}+\frac{2}{17.19}+\frac{4}{19.23}+\frac{2}{23.25}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{5}{25}-\frac{1}{25}\)
\(=\frac{4}{25}\)
P/S: Dấu "." là dấu nhân bạn nhé
a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
\(\dfrac{13+x}{20}\) = \(\dfrac{3}{4}\)
13 + \(x\) = 20 \(\times\) \(\dfrac{3}{4}\)
13 + \(x\) = 15
\(x\) = 15 - 13
\(x\) = 2
Cách khác :
\(\dfrac{13+x}{20}=\dfrac{3}{4}\)
\(\dfrac{13+x}{20}=\dfrac{15}{20}\)
\(13+x=15\)
\(x=15-13\)
\(x=2\)
A = \(\dfrac{2}{35}\) + \(\dfrac{4}{77}\) + \(\dfrac{2}{143}\) + \(\dfrac{4}{221}\) + \(\dfrac{2}{323}\) + \(\dfrac{4}{437}\) + \(\dfrac{2}{575}\)
A = \(\dfrac{2}{5\times7}\)+\(\dfrac{4}{7\times11}\)+\(\dfrac{2}{11\times13}\)+\(\dfrac{4}{13\times17}\)+\(\dfrac{2}{17\times19}\)+\(\dfrac{4}{19\times23}\)+\(\dfrac{2}{23\times25}\)
A = \(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{13}\)+\(\dfrac{1}{13}\)-\(\dfrac{1}{17}\)+\(\dfrac{1}{17}\)-\(\dfrac{1}{19}\)+\(\dfrac{1}{19}\)-\(\dfrac{1}{23}\)+\(\dfrac{1}{23}\)-\(\dfrac{1}{25}\)
A = \(\dfrac{1}{5}\) - \(\dfrac{1}{25}\)
A = \(\dfrac{4}{25}\)