Question

giúp mk câu 1 và 3

Answer 1

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={\(-\dfrac{1}{2}\)}

Answer 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: \(S=\left\{0;-\dfrac{1}{2}\right\}\)