Lời giải:
\((2x-1)^4=2(x^2-x+1)\)
\(\Leftrightarrow [(2x-1)^2]^2=2(x^2-x+1)\)
\(\Leftrightarrow (4x^2-4x+1)^2=2(x^2-x+1)\)
Đặt \(x^2-x=a\) thì PT trở thành:
\((4a+1)^2=2(a+1)\)
\(\Leftrightarrow 16a^2+6a-1=0\)
\(\Leftrightarrow (8a-1)(2a+1)=0\Rightarrow \left[\begin{matrix} a=\frac{1}{8}\\ a=-\frac{1}{2}\end{matrix}\right.\)
Nếu \(a=\frac{1}{8}\Leftrightarrow x^2-x=\frac{1}{8}\)
\(\Leftrightarrow (x-\frac{1}{2})^2=\frac{3}{8}\Rightarrow x=\pm \sqrt{\frac{3}{8}}+\frac{1}{2}\) (chọn)
Nếu \(a=-\frac{1}{2}\Leftrightarrow x^2-x=-\frac{1}{2}\Leftrightarrow (x-\frac{1}{2})^2=-\frac{1}{4}< 0\) (loại)
Vậy \(x=\pm \sqrt{\frac{3}{8}}+\frac{1}{2}\)
