[tex] \frac{x+1}{2}+ \frac{x+2}{3}+...+ \frac{x+99}{100}= \frac{x-1+2}{2}+ \frac{x-1+3}{3}+...+ \frac{x-1+100}{100}= [/tex].[tex]\frac{x-1}{2}+1+ \frac{x-1}{3}+1+...+ \frac{x-1}{100}+1[/tex],se,obtine,[tex](x-1) (\frac{1}{2}+ \frac{1}{3}+...+ \frac{1}{100})+99=99 [/tex].
Reducem 99 din stanga cu cel din dreaptasi se obtinem :
[tex](x-1)( \frac{1}{2}+ \frac{1}{3}+...+ \frac{1}{100})=0 [/tex], a doua paranteza e ≠ 0, deci x-1=0, de unde x=1.
