Salut,
[tex]Membrul\;drept:\\\\
\frac{x-1}{x}+\frac{x-2}{x}+\ldots+\frac{1}{x}=\frac{x-1}{x}+\frac{x-2}{x}+\ldots+\frac{x-(x-1)}{x},\;deci\;membrul\\\\drept\;con\c{t}ine\;x-1\;frac\c{t}ii\;(1).\\\\\frac{x-1}{x}+\frac{x-2}{x}+\ldots+\frac{x-(x-1)}{x}=1-\frac{1}{x}+1-\frac{2}{x}+\ldots+1-\frac{x-1}{x}=\\\\=\underbrace{1+1+\ldots+1}_{de\;x-1\;ori,\;vezi\;(1)}-\left(\frac{1}{x}+\frac{2}{x}+\ldots+\frac{x-1}{x}\right)=1\cdot(x-1)-\frac{1}{x}\cdot(1+2+\ldots+x-1)=\\\\\\=x-1-\frac{1}{x}\cdot\frac{(1+x-1)\cdot(x-1)}{2}=\ldots=\frac{x-1}{2}=3,\;deci\;x-1=6,\\\\deci\;x=7,\;asta\;este\;solu\c{t}ia.[/tex]
Green eyes.
