\(x^2\left(x-1\right)^2=\left(2x-1\right)^2+2\)
\(\Leftrightarrow x^2\left(x-1\right)^2=\left[x+\left(x-1\right)\right]^2+2\)
\(\Leftrightarrow x^2\left(x-1\right)^2=4x^2-4x+1+2\)
\(\Leftrightarrow x^2\left(x-1\right)^2-4x\left(x-1\right)-3=0\) (1)
Đặt \(a=x\left(x-1\right)\)
\(x\left(x-1\right)=x^2-x=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(ĐK:a^2\ge2\Leftrightarrow\left|a\right|\ge\sqrt{2}\)
\(\left(1\right)\Leftrightarrow a^2-4a-3=0\)
\(\Delta=\left(-4\right)^2-4.\left(-3\right)=16+12=28>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{4+\sqrt{28}}{2}=2+\sqrt{7}\left(tm\right)\\a=\dfrac{4-\sqrt{28}}{2}=2-\sqrt{7}\left(ktm\right)\end{matrix}\right.\)
\(\rightarrow x\left(x-1\right)=2+\sqrt{7}\)
\(\Leftrightarrow x^2-x-\left(2+\sqrt{7}\right)=0\)
\(\Leftrightarrow x=\dfrac{1\pm\sqrt{9+4\sqrt{7}}}{2}\)
Vậy \(S=\left\{\dfrac{1\pm\sqrt{9+4\sqrt{7}}}{2}\right\}\)
