Đáp án:

\(S = \left\{-\dfrac54;1\right\}\)

Giải thích các bước giải:

\(\begin{array}{l}
\quad \dfrac{113}{64} - 2\sqrt{x^2 + \dfrac x4 - 1} = \dfrac12(x-1) + \left(x - \dfrac18\right)^2\qquad (*)\\
ĐK:x^2 + \dfrac x4 - 1\geqslant 0\Leftrightarrow \left[\begin{array}{l}x \geqslant \dfrac{-1 + \sqrt{65}}{8}\\x \leqslant \dfrac{-1-\sqrt{65}}{8}\end{array}\right.\\
(*)\Leftrightarrow \dfrac{113}{64} - 2\sqrt{x^2 + \dfrac x4 - 1} = \dfrac x2 - \dfrac12 + x^2 - \dfrac x4 + \dfrac{1}{64}\\
\Leftrightarrow x^2 + \dfrac x4 +2\sqrt{x^2 + \dfrac x4 - 1}  - \dfrac94 =0\\
\text{Đặt}\ t = \sqrt{x^2 + \dfrac x4 - 1},\quad t\geqslant 0\\
\Rightarrow t^2 = x^2 + \dfrac x4 - 1\\
\text{Phương trình trở thành:}\\
\quad t^2 + 1 + 2t - \dfrac94 = 0\\
\Leftrightarrow 4t^2 + 8t - 5 =0\\
\Leftrightarrow (2t - 1)(2t + 5) = 0\\
\Leftrightarrow \left[\begin{array}{l}t = \dfrac12\qquad (n)\\t = - \dfrac52\quad (l)\end{array}\right.\\
\text{Với}\ t = \dfrac12\ \text{ta được:}\\
\quad \sqrt{x^2 + \dfrac x4 - 1} = \dfrac12\\
\Rightarrow x^2 + \dfrac x4 - 1 = \dfrac14\\
\Leftrightarrow 4x^2 + x - 5 = 0\\
\Leftrightarrow (x - 1)(4x + 5) = 0\\
\Leftrightarrow \left[\begin{array}{l}x= 1\\x = - \dfrac54\end{array}\right.\quad \text{(nhận)}\\
\text{Vậy}\ S = \left\{-\dfrac54;1\right\}
\end{array}\)