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a ) \(x+7\frac{4}{5}=9\frac{7}{15}\)
\(x+\frac{39}{5}=\frac{142}{15}\)
\(x=\frac{142}{15}-\frac{39}{5}\)
\(x=\frac{5}{3}\)
b ) \(x-4\frac{9}{11}=\frac{2121}{2222}\)
\(x-\frac{53}{11}=\frac{21}{22}\)
\(x=\frac{21}{22}+\frac{53}{11}\)
\(x=\frac{127}{22}\)
\(x+7\frac{4}{5}=9\frac{7}{15}\)
\(x=9\frac{7}{15}-7\frac{4}{5}\)
\(x=9\frac{7}{15}-7\frac{12}{15}\)
\(x=-2\frac{1}{3}\)
\(x-4\frac{9}{11}=\frac{2121}{2222}\)
\(x-4\frac{9}{11}=\frac{101.21}{101.22}\)
\(x-\frac{53}{11}=\frac{21}{22}\)
\(x=\frac{21}{22}+\frac{53}{11}\)
\(x=\frac{127}{22}=5\frac{17}{22}\)
=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; (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
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
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
\(71+52,5\times4=\frac{x+140}{x}+210\)
\(71+210=\frac{x+140}{x}+210\)
\(=>\frac{x+140}{x}=71\)
\(71=\frac{142}{2}\)\(\Rightarrow x=142-140=2\)
\(x+7\dfrac{4}{5}=9\dfrac{7}{15}\)
\(=>x+\dfrac{39}{5}=\dfrac{142}{15}\)
\(=>x=\dfrac{142}{15}-\dfrac{39}{5}=\dfrac{142}{15}-\dfrac{117}{15}\)
\(=>x=\dfrac{25}{15}=\dfrac{5}{3}\)
___________
\(x-4\dfrac{9}{11}=\dfrac{2121}{2222}\)
\(=>x-\dfrac{53}{11}=\dfrac{21}{22}\)
\(=>x=\dfrac{21}{22}+\dfrac{53}{11}=\dfrac{21}{22}+\dfrac{106}{22}\)
\(=>x=\dfrac{127}{22}\)
\(x+7\dfrac{4}{5}=9\dfrac{7}{15}\)
\(x=9\dfrac{7}{15}-7\dfrac{4}{5}\)
\(x=\left(9-7\right)+\left(\dfrac{7}{15}-\dfrac{4}{5}\right)\)
\(x=2\dfrac{-1}{3}=\dfrac{5}{3}\)
\(x=\dfrac{2121:101}{2222:101}+4\dfrac{9}{11}\)
\(x=\dfrac{21}{22}+4\dfrac{18}{22}\)
\(x=\dfrac{21+4\cdot22+18}{22}=\dfrac{127}{22}\)