\(\Leftrightarrow\left\{{}\begin{matrix}x^2-\dfrac{1}{2}x=0\\x^2-\dfrac{1}{4}=0\end{matrix}\right.\) ( vì giá trị tuyệt đối ko âm )
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x-\dfrac{1}{2}\right)=0\\x^2=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(S=\left\{0;\dfrac{1}{2}\right\}\)
Dấu ''='' xảy ra khi
\(\left\{{}\begin{matrix}x\left(x-\dfrac{1}{2}\right)=0\\\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0orx=\dfrac{1}{2}\\x=\dfrac{1}{2}orx=-\dfrac{1}{2}\end{matrix}\right.\)
Vì \(\left|x^2+\dfrac{1}{2}x\right|\ge0;\left|x^2-\dfrac{1}{4}\right|\ge0\Rightarrow\left|x^2+\dfrac{1}{2}x\right|+\left|x^2-\dfrac{1}{4}\right|\ge0\)
Mà \(\left|x^2+\dfrac{1}{2}x\right|+\left|x^2-\dfrac{1}{4}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+\dfrac{1}{2}x=0\\x^2-\dfrac{1}{4}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+\dfrac{1}{2}\right)=0\\x^2=\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\x=\pm\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=-\dfrac{1}{2}\)
Vì \(\left|x^2-\dfrac{1}{2}x\right|\ge0;\left|x^2-\dfrac{1}{4}\right|\ge0\Rightarrow\left|x^2-\dfrac{1}{2}x\right|+\left|x^2-\dfrac{1}{4}\right|\ge0\)
Mà \(\left|x^2-\dfrac{1}{2}x\right|+\left|x^2-\dfrac{1}{4}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x^2-\dfrac{1}{2}x=0\\x^2-\dfrac{1}{4}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x-\dfrac{1}{2}\right)=0\\x^2=\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\\x=\pm\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{1}{2}\)