ĐK: x>0
\(\sqrt{\dfrac{1}{x}}=\dfrac{5}{2}-8x^2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{2}-8x^2\ge0\\\dfrac{1}{x}=\left(\dfrac{5}{2}-8x^2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-\sqrt{5}}{4}\le x\le\dfrac{\sqrt{5}}{4}\\\dfrac{1}{x}=\dfrac{25}{4}-40x^2+64x^4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-\sqrt{5}}{4}\le x\le\dfrac{\sqrt{5}}{4}\\64x^5-40x^3+\dfrac{25}{4}x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-\sqrt{5}}{4}\le x\le\dfrac{\sqrt{5}}{4}\\\left(4x-1\right)^2\left(4x^3+2x^2-\dfrac{7}{4}x-1\right)=0\end{matrix}\right.\)
pt \(4x^3+2x^2-\dfrac{7}{4}x-1\) có 1 nghiệm \(x\approx0.681144>\dfrac{\sqrt{5}}{4}\) (loại)
Vậy \(S=\left\{\dfrac{1}{4}\right\}\)
