\(a) 2x+\dfrac{4}{3}=0\\↔2x=-\dfrac{4}{3}\\↔x=-\dfrac{2}{3}\\Vậy\,\,x=-\dfrac{2}{3}\\b) (2+x)(8-6x)=0\\↔\left[\begin{array}{1}2+x=0\\8-6x=0\end{array}\right.\\↔\left[\begin{array}{1}x=-2\\6x=8\end{array}\right.\\↔\left[\begin{array}{1}x=-2\\x=\dfrac{4}{3}\end{array}\right.\\Vậy\,\,x=\{-2;\dfrac{4}{3}\}\\c)x^3+9x=0\\↔x(x^2+9)=0\\↔\left[\begin{array}{1}x=0\\x^2+9=0\end{array}\right.\\↔\left[\begin{array}{1}x=0\\x^2=-9(vô\,\,lý)\end{array}\right.\\↔x=0\\Vậy\,\,x=0\)
