[tex]\boxed{\text{Conditie de existenta: }x \neq 0}[/tex]
[tex]
\dfrac{x-2}{x} + \dfrac{x-4}{x} + \dfrac{x-6}{x} + ...+\dfrac{2}{x} = 12\ \ \\ \dfrac{x-2}{x} + \dfrac{x-4}{x} + \dfrac{x-6}{x} + ...+\dfrac{x-(x-2)}{x} = 12 \\ \\ \sum\limits_{k=1}^{\frac{x-2}{2}} \dfrac{x-2\cdot k}{x} = 12 \\ \sum\limits_{k=1}^{\frac{x-2}{2}}\Big(\dfrac{x}{x}-\dfrac{2\cdot k}{x}\Big)=12 \\ \sum\limits_{k=1}^{\frac{x-2}{2}}\Big(1-\dfrac{2}{x}\cdot k\Big)=12 [/tex]
[tex]\sum\limits_{k=1}^{\frac{x-2}{2}}1 - \dfrac{2}{x}\sum\limits_{k=1}^{\frac{x-2}{2}}k = 12 \\ \\ \dfrac{x-2}{2}-\dfrac{2}{x}\cdot \dfrac{\dfrac{x-2}{2}\cdot \Big(\dfrac{x-2}{2}+1\Big)}{2} = 12 \\ \\\dfrac{x-2}{2}-\dfrac{1}{x}\cdot \dfrac{x-2}{2}\cdot \Big(\dfrac{x-2}{2}+1\Big) \\ \\ \dfrac{x-2}{2}\cdot \left(1-\dfrac{1}{x}\cdot \Big(\dfrac{x-2}{2}+1\Big)\right) = 12\\ \\ \dfrac{x-2}{2}\cdot \left(1-\dfrac{1}{x}\cdot \Big(\dfrac{x-2+2}{2}\Big)\right) = 12[/tex]
[tex]\dfrac{x-2}{2}\cdot \Big(1-\dfrac{1}{x}\cdot \dfrac{x}{2}\Big) = 12 \\ \\ \dfrac{x-2}{2}\cdot \Big(1-\dfrac{1}{2}\Big) = 12 \\ \\ \dfrac{x-2}{2}\cdot \dfrac{1}{2} = 12\\ \\ \dfrac{x-2}{4} = 12 \\ \\ x-2 = 12\cdot 4 \\ \\ x- 2 = 48 \\ \\ \boxed{\boxed{x = 50}}[/tex]
