Question

X+1 supra 2 + x+2 supra 3+x+3supra4+...+x+99supra 100=99.Va rog sa ma ajutati!:))

Answer 1

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