Giải phương tình: \(x+\sqrt{2x-1}=2\left(x-3\right)^2\)

Điều kiện: \(x\ge\dfrac{1}{2}\)

\(PT\Leftrightarrow\sqrt{2x-1}-3=2x^2-13x+15\\ \Leftrightarrow\dfrac{2x-10}{\sqrt{2x-1}-3}=\left(x-5\right)\left(2x-3\right)\\ \Leftrightarrow\left(x-5\right)\left(\dfrac{2}{\sqrt{2x-1}+3}-2x+3\right)=0\\ \Leftrightarrow\begin{matrix}x=5\\\dfrac{2}{\sqrt{2x-1}+3}=2x-3\left(1\right)\end{matrix}\)

\(\left(1\right)\Leftrightarrow\left(2x-3\right)\left(\sqrt{2x-1}+3\right)=2\)

Đặt \(t=\sqrt{2x-1},t>0\) phương trình trở thành \(\left(t^2-2\right)\left(t+3\right)=2\\ \)

\(\Leftrightarrow\left[{}\begin{matrix}t=-2\left(L\right)\\t=\dfrac{-1-\sqrt{17}}{2}\\t=\dfrac{-1+\sqrt{17}}{2}\end{matrix}\right.\left(L\right)\)

Với \(t=\dfrac{-1+\sqrt{17}}{2}\) ta có \(\sqrt{2x-1}=\dfrac{-1+\sqrt{17}}{2}\)

\(\Leftrightarrow2x-1=\dfrac{9-\sqrt{17}}{2}\)

\(\Leftrightarrow x=\dfrac{11-\sqrt{17}}{4}\)

Vậy \(E=\left\{5;\dfrac{11-\sqrt{17}}{4}\right\}\)