Giải thích các bước giải:
`(3x+1)^2=-9`
`=>(3x+1)^2=(+-3i)^2`
`=>`\(\left[ \begin{array}{l}3x+1=3i\\3x+1=-3i\end{array} \right.\) `=>`\(\left[ \begin{array}{l}3x=-1+3i\\3x=-1-3i\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\dfrac{-1+3i}3=-\dfrac13+i\\x=\dfrac{-1-3i}3=-\dfrac13-i\end{array} \right.\)
Vậy `x in{-1/3+i;-1/3-i}.`