a,
Ta có: $\frac {3x-1}{\frac {-1}{2x}+5}=0$
<=> $ \left[ \begin{array}{l}3x-1=0\\\frac {-1}{2x}+5=0\end{array} \right. $
<=>$\left[ \begin{array}{l}x=\frac{1}{3}\\x=\frac{1}{10}\end{array} \right. $
Vậy $\left[ \begin{array}{l}x=\frac{1}{3}\\x=\frac{1}{10}\end{array} \right. $
b,
Ta có: $\frac {1}{2x} + \frac {2}{3x} -1= \frac {-8}{3}$
<=> $\frac {7x}{6x^{2}}-1= \frac {-8}{3}$
<=> $\frac {7x-6x^{2}}{6x^{2}}= \frac {-8}{3}$
<=> $21x-18x^{2}=-48x^{2}$
<=> $30x^{2}+21x=0$
<=> $\frac {-30}{21} = \frac {x}{x^{2}}$
<=> $x=-0,7$.
Vậy $x=-0,7$
