\(1+3+5+...+99=\dfrac{\left(\dfrac{99-1}{2}+1\right)\cdot\left(99+1\right)}{2}=\dfrac{50\cdot100}{2}=\dfrac{5000}{2}=2500=\left(x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=50\\x-2=-50\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=52\\x=-48\end{matrix}\right.\)
Vậy \(x=52\text{ hoặc }x=-48\)