Question

(9x^2-1)^2|x-1/3|=0

Answer 1

\(\left(9x^2-1\right)^2\left|x-\dfrac{1}{3}\right|=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9x^2-1=0\\x-\dfrac{1}{3}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{1}{9}\\x=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy...

Answer 2

Ta có: \(\left(9x^2-1\right)^2\cdot\left|x-\dfrac{1}{3}\right|=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9x^2-1=0\\x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)

Answer 3

\(\left(9x^2-1\right)^2.\left|x-\dfrac{1}{3}\right|=0\) 

\(\Rightarrow\left[{}\begin{matrix}\left(9x^2-1\right)^2=0\\\left|x-\dfrac{1}{3}\right|=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}9x^2-1=0\\x-\dfrac{1}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{1}{3}\end{matrix}\right.\) 

Vậy \(x\in\left\{\dfrac{-1}{3};\dfrac{1}{3}\right\}\)