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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 = \(\dfrac{502}{3}\)
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
\(=\dfrac{29}{13}\cdot\dfrac{13}{29}\cdot4\cdot\dfrac{21}{8}\cdot\dfrac{8}{21}=4\)
a; A = \(\dfrac{4026\times2014+4030}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2014+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2013\times2+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-4026+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2011\right)}{2013\times2016-2011}\)
A = 2
\(\dfrac{15}{14}\): \(\dfrac{10}{21}\) \(\times\) \(\dfrac{1}{5}\) = \(\dfrac{15}{14}\) \(\times\) \(\dfrac{21}{10}\) \(\times\) \(\dfrac{1}{5}\) = \(\dfrac{5\times3\times7\times3}{7\times2\times10\times5}\) = \(\dfrac{9}{20}\)
5 \(\times\) \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) = 1 + \(\dfrac{1}{5}\) = \(\dfrac{6}{5}\)
7 : \(\dfrac{1}{5}\) - \(\dfrac{1}{5}\) = 35 - \(\dfrac{1}{5}\) = \(\dfrac{174}{5}\)
6 + \(\dfrac{1}{5}\): 2 = 6 + \(\dfrac{1}{10}\) = \(\dfrac{61}{10}\)
8 - \(\dfrac{1}{5}\) \(\times\) 7 = 8 - \(\dfrac{7}{5}\) = \(\dfrac{33}{5}\)
\(\dfrac{15}{14}\) : \(\dfrac{10}{21}\) x \(\dfrac{1}{5}\) = \(\dfrac{15}{14}\) x \(\dfrac{21}{10}\) x \(\dfrac{1}{5}\) = \(\dfrac{9}{4}\) x \(\dfrac{1}{5}\) = \(\dfrac{9}{20}\)
5 x \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) = \(\dfrac{5}{1}\) x \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) = 1 x \(\dfrac{1}{5}\) = \(\dfrac{1}{5}\)
7 : \(\dfrac{1}{5}-\dfrac{1}{5}\) = \(\dfrac{7}{1}\) x \(\dfrac{5}{1}-\dfrac{1}{5}\) = \(\dfrac{35}{1}\) - \(\dfrac{1}{5}\) = \(\dfrac{175}{5}\) - \(\dfrac{1}{5}\) = \(\dfrac{174}{5}\)
6 + \(\dfrac{1}{5}\) : 2 = \(\dfrac{6}{1}\) + \(\dfrac{1}{5}\) x \(\dfrac{1}{2}\) = \(\dfrac{6}{1}+\dfrac{1}{10}\) = \(\dfrac{60}{10}\) + \(\dfrac{1}{10}\) = \(\dfrac{61}{10}\)
8 - \(\dfrac{1}{5}\) x 7 = \(\dfrac{8}{1}\) - \(\dfrac{1}{5}\) x \(\dfrac{7}{1}\) = \(\dfrac{8}{1}-\dfrac{7}{5}\) = \(\dfrac{40}{5}\) - \(\dfrac{7}{5}\) = \(\dfrac{33}{5}\)
Sai Báo Lại Mình Nha!
\(\dfrac{13+x}{20}\) = \(\dfrac{3}{4}\)
13 + \(x\) = 20 \(\times\) \(\dfrac{3}{4}\)
13 + \(x\) = 15
\(x\) = 15 - 13
\(x\) = 2
Cách khác :
\(\dfrac{13+x}{20}=\dfrac{3}{4}\)
\(\dfrac{13+x}{20}=\dfrac{15}{20}\)
\(13+x=15\)
\(x=15-13\)
\(x=2\)
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