\(10+\left(2x-1\right)^2:3=13\)
\(\left(2x-1\right)^2:3=3\)
\(\left(2x-1\right)^2=9\)
\(\Rightarrow\left(2x-1\right)=\sqrt{9}\) hoặc \(2x-1=-\sqrt{9}\)
TH1 \(2x-1=\sqrt{9}\)
\(2x-1=3\)
\(2x=4\)
\(x=2\)
TH2 \(2x-1=-\sqrt{9}\)
\(2x-1=-3\)
\(2x=-2\)
\(x=-1\)
Vậy \(x=2\) hoặc \(x=-1\)