\(\Leftrightarrow\left[\left(99-1\right)\div2+1\right]\div2\times\left(99+1\right)=\left(x-2\right)^2\)
\(\Leftrightarrow\left[98\div2+1\right]\div2\times98=\left(x-2\right)^2\)
\(\Leftrightarrow\frac{50\times98}{2}=\left(x-2\right)^2\)
\(\Leftrightarrow2450=\left(x-2\right)^2\)
\(\Leftrightarrow\sqrt{2450}=\left(x-2\right)\)
\(\Leftrightarrow x=\sqrt{2450}+2\)
Ta có:
\(1+3+5+...+99=100.\frac{50}{2}=2500\)
\(\Rightarrow\left(x-2\right)^2=2500\)
\(\Rightarrow\left[{}\begin{matrix}x-2=50\\x-2=-50\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=52\left(TM\right)\\x=-48\left(\text{ loại}\right)\end{matrix}\right.\)
Vậy \(x=52\) thỏa mãn đề bài
