`a)`
`(3x^2-1/3)(x^3+1/8)=0`
`=>`\(\left[ \begin{array}{l}3x^2-\dfrac{1}{3}=0\\x^3+\dfrac{1}{8}=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `x∈{1/3;-1/3;1/2}`
`b)`
`(|x|-1/2)(4x^2+1)=0`
`=>`\(\left[ \begin{array}{l}|x|-\dfrac{1}{2}=0\\4x^2+1=0\end{array} \right.\)
Xét `4x^2+1=0`
`=>4x^2=-1`
Vì `4x^2>=0∀x`
Mà `4x^2=-1` (vô lí)
`=>|x|=1/2`
`=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `x∈{1/2;-1/2}`
`c)`
`(3x^3+1/9)(|x|+3)=0`
`=>`\(\left[ \begin{array}{l}3x^3+\dfrac{1}{9}=0\\|x|+3=0\end{array} \right.\)
Xét `|x|+3=0`
`=>|x|=-3`
Vì `|x|>=0∀x`
Mà `|x|=-3` (vô lí)
`=>3x^3+1/9=0`
`=>3x^3=-1/9`
`=>x^3=-1/27`
`=>x=-1/3`
Vậy `x=-1/3`
