1. \(\left(3x+\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
⇔\(3x+\dfrac{1}{2}=x-\dfrac{1}{2}\)
⇔\(3x-x=-\dfrac{1}{2}-\dfrac{1}{2}\)
⇔\(2x=-1\)
⇔\(x=-\dfrac{1}{2}\)
Vậy S={\(-\dfrac{1}{2}\)}
1) Ta có: \(\left(3x+\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow\left(3x+\dfrac{1}{2}\right)^2-\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(3x+\dfrac{1}{2}-x+\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}+x-\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-\dfrac{1}{2}\right\}\)