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a) x/2012 + (x+1)/2013 + (x+2)/2014 + (x+3)/2015 + (x+4)/2016 = 5
b) (x-90)/10 + (x-76)/12 + (x-58)/14 + (x-36)/16 + (x-15)/17 = 15
c) (x-3)/2011 + (x-2)/2012 = (x-2012)/2 + (x-2011)/3
d) (x+2)/2008 + (x+3)/2007 + (x+4)/2006 + (x+2028)/6 = 0

 

[Blue Plus]

x/2012 + (x+1)/2013 + (x+2)/2014 + (x+3)/2015 + (x+4)/2016 = 5
b) (x-90)/10 + (x-76)/12 + (x-58)/14 + (x-36)/16 + (x-15)/17 = 15
c) (x-3)/2011 + (x-2)/2012 = (x-2012)/2 + (x-2011)/3
d) (x+2)/2008 + (x+3)/2007 + (x+4)/2006 + (x+2028)/6 = 0

$ \frac{x}{2012} + \frac{x + 1}{2013} + \frac{x + 2}{2014} + \frac{x + 3}{2015} + \frac{x + 4}{2016} = 5 \\ \Leftrightarrow \frac{x}{2012} - 1 + \frac{x + 1}{2013} - 1 + \frac{x + 2}{2014} - 1 + \frac{x + 3}{2015} - 1 + \frac{x + 4}{2016} - 1 = 0 \\ \Leftrightarrow \frac{x - 2012}{2012} + \frac{x - 2012}{2013} + \frac{x - 2012}{2014} + \frac{x - 2012}{2015} + \frac{x - 2012}{2016} = 0 \\ \Leftrightarrow (x - 2012)\left ( \frac{1}{2012} + \frac{1}{2013} + \frac{1}{2014} + \frac{1}{2015} + \frac{1}{2016}\right ) = 0 \\ \frac{1}{2012} + \frac{1}{2013} + \frac{1}{2014} + \frac{1}{2015} + \frac{1}{2016} \neq 0 \Rightarrow x - 2012 = 0 \Leftrightarrow x = 2012 $
$ \frac{x - 90}{10} + \frac{x - 76}{12} + \frac{x - 58}{14} + \frac{x - 36}{16} + \frac{x - 15}{17} = 15 \\ \Leftrightarrow \frac{x - 90}{10} - 1 + \frac{x - 76}{12} - 2 + \frac{x - 58}{14} - 3 + \frac{x - 36}{16} - 4 + \frac{x - 15}{17} - 5 = 0 \\ \Leftrightarrow \frac{x - 100}{10} + \frac{x - 100}{12} + \frac{x - 100}{14} + \frac{x - 100}{16} + \frac{x - 100}{17} = 0 \\ \Leftrightarrow (x - 100)\left ( \frac{1}{10} + \frac{1}{12} + \frac{1}{14} + \frac{1}{16} + \frac{1}{17}\right ) = 0 \\ \frac{1}{10} + \frac{1}{12} + \frac{1}{14} + \frac{1}{16} + \frac{1}{17} \neq 0 \Rightarrow x - 100 = 0 \Leftrightarrow x = 100 $
$ \frac{x - 3}{2011} + \frac{x - 2}{2012} = \frac{x - 2012}{2} + \frac{x - 2011}{3} \\ \Leftrightarrow \frac{x - 3}{2011} - 1 + \frac{x - 2}{2012} - 1 = \frac{x - 2012}{2} - 1 + \frac{x - 2012}{3} - 1 \\ \Leftrightarrow \frac{x - 2014}{2011} + \frac{x - 2014}{2012} = \frac{x - 2014}{2} + \frac{x - 2014}{3} \\ \Leftrightarrow \frac{x - 2014}{2011} + \frac{x - 2014}{2012} - \frac{x - 2014}{2} - \frac{x - 2014}{3} = 0 \\\Leftrightarrow (x - 100)\left ( \frac{1}{2011} + \frac{1}{2012} - \frac{1}{2} - \frac{1}{3} \right) = 0 \\ \frac{1}{2011} + \frac{1}{2012} - \frac{1}{2} - \frac{1}{3} \neq 0 \Rightarrow x - 2014 = 0 \Leftrightarrow x = 2014 $
$ \frac{x + 2}{2008} + \frac{x + 3}{2007} + \frac{x + 4}{2006} + \frac{x + 2028}{6} = 0 \\ \Leftrightarrow \frac{x + 2}{2008} + 1 + \frac{x + 3}{2007} + 1 + \frac{x + 4}{2006} + 1 + \frac{x + 2028}{6} - 3 = 0 \\ \Leftrightarrow \frac{x + 2010}{2008} + \frac{x + 2010}{2007} + \frac{x + 2010}{2006} + \frac{x + 2010}{6} = 0 \\ \Leftrightarrow (x + 2010)\left ( \frac{1}{2008} + \frac{1}{2007} + \frac{1}{2006} + \frac{1}{6} \right ) = 0 \\ \frac{1}{2008} + \frac{1}{2007} + \frac{1}{2006} + \frac{1}{6} \neq 0 \Rightarrow x + 2010 = 0 \Leftrightarrow x = - 2010 $

 

[hophuonganh0207]

[tex]\fn_phv (\frac{x}{2012}-1) + (\frac{x+1}{2013}-1) + (\frac{x+2}{2014}-1) + (\frac{x+3}{2015}-1) + (\frac{x+4}{2016}-1)= 0 => \frac{x-2012}{2012}+ \frac{x-2012}{2013}+ \frac{x-2012}{2014}+ \frac{x-2012}{2015}+ \frac{x-2012}{2016}= 0 => (x-2012)*(\frac{1}{2012}+ \frac{1}{2013}+ \frac{1}{2014}+ \frac{1}{2015}+ \frac{1}{2016})= 0 do (1/2012 + 1/2013+...+1/2016) khác 0 dấu '=' xảy ra <=> x-2012= 0 \rightarrow x= 2012[/tex]