Question

Rezolvati ecuatia :
$\frac{x+1}{2} + \frac{x+2}{3} + \frac{x+3}{4} +...+ \frac{x+2012}{2013} =2012$.
EU CONSIDER CA SINGURA SOLUTIE E X=1.... voi ce ziceti ?

Answer 1

$\frac{x+k}{k+1}-1= \frac{x-1}{k+1} \\ \\ \\ \frac{x+1}{2}+ \frac{x+2}{3}+...+ \frac{x+2012}{2013} =1+1+...+1~(2012~de~1) \\ \\ ( \frac{x+1}{2}-1) +(\frac{x+2}{3}-1)+...+( \frac{x+2012}{2013}-1)=0 \\ \\ \frac{x-1}{2}+ \frac{x-1}{3} +...+ \frac{x-1}{2013} =0 \\ \\ (x-1)( \frac{1}{2}+ \frac{1}{3}+...+ \frac{1}{2013} )=0 \Rightarrow x-1=0 \Rightarrow x=1.$