Đáp án:
$\begin{array}{l}
a)\frac{{3{x^2} + 7x - 10}}{x} = 0\left( {dkxd:x \ne 0} \right)\\
\Rightarrow 3{x^2} + 7x - 10 = 0\\
\Rightarrow 3{x^2} - 3x + 10x - 10 = 0\\
\Rightarrow \left( {x - 1} \right)\left( {3x + 10} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x = - \frac{{10}}{3}\left( {tm} \right)
\end{array} \right.\\
b)\frac{{4x - 17}}{{2{x^2} + 1}} = 0\left( {Dkxd:\forall x} \right)\\
\Rightarrow 4x - 17 = 0\\
\Rightarrow x = \frac{{17}}{4}\\
c)\frac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {dkxd:x \ne 3} \right)\\
\Rightarrow {x^2} - x - 6 = 0\\
\Rightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\left( {ktm} \right)\\
x = - 2\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = - 2
\end{array}$
