Ta có: \(1+3+5+...+\left(2x-1\right)=1069\)
\(\Leftrightarrow\frac{\left(2x-1+1\right)\cdot\left[\left(2x-1-1\right)\div2+1\right]}{2}=1069\)
\(\Leftrightarrow\frac{2x\cdot\left(x-1+1\right)}{2}=1069\)
\(\Leftrightarrow x^2=1069\)
\(\Rightarrow x=\sqrt{1069}\)
Số số hạng
\(\left(2x-1-1\right):2+1=x\)
Tổng
\(\left(2x-1+1\right)\cdot x:2=x^2\)
\(x^2=1069\left(x\ge0\right)\)
\(x=\sqrt{1069}\)
\(x=33\)