Đáp án:
$\begin{array}{l}
a)\frac{{98{x^2} - 2}}{{x - 2}} = 0\left( {dkxd:x \ne 2} \right)\\
\Rightarrow 98{x^2} - 2 = 0\\
\Rightarrow {x^2} = \frac{1}{{49}}\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{1}{7}\\
x = - \frac{1}{7}
\end{array} \right.\\
b)\frac{{3x - 2}}{{{x^2} + 2x + 1}} = 0\left( {dkxd:x \ne - 1} \right)\\
\Rightarrow 3x - 2 = 0\\
\Rightarrow x = \frac{2}{3}\left( {tmdk} \right)\\
Vậy\,x = \frac{2}{3}
\end{array}$
