Giải thích các bước giải:
Ta có: $\dfrac{1 - \sqrt{1 - 4x^{2}}}{x} < 3$
$\to \dfrac{1 - \sqrt{1 - 4x^{2}}}{x} - 3 < 0$
$\to \dfrac{1 - \sqrt{1 - 4x^{2}} - 3x}{x} < 0$
*TH1:
$\left\{\begin{matrix}1 - \sqrt{1 - 4x^{2}} - 3x > 0\\ x < 0\end{matrix}\right.$
$\to \left\{\begin{matrix}\sqrt{1 - 4x^{2}} < 1 - 3x\\ x < 0\end{matrix}\right.$
$\to \left\{\begin{matrix}1 - 4x^{2} \geq 0\\1 - 3x > 0\\1 - 4x^{2} < \left ( 1 - 3x \right )^{2}\\ x < 0\end{matrix}\right.$
$\to \left\{\begin{matrix}4x^{2} \leq 1\\3x < 1\\1 - 4x^{2} < 1 - 6x + 9x^{2}\\ x < 0\end{matrix}\right.$
$\to \left\{\begin{matrix}x^{2} \leq \dfrac{1}{4}\\x < \dfrac{1}{3}\\13x^{2} - 6x > 0\\ x < 0\end{matrix}\right.$
$\to -\dfrac{1}{2} \leq x < 0$
TH2:
$\left\{\begin{matrix}1 - \sqrt{1 - 4x^{2}} - 3x < 0\\ x > 0\end{matrix}\right.$
$\to \left\{\begin{matrix}\sqrt{1 - 4x^{2}} > 1 - 3x \left ( 1 \right )\\ x > 0\end{matrix}\right.$
Giải $\left ( 1 \right )$:
+) $\left\{\begin{matrix}1 - 3x < 0\\ 1 - 4x^{2} \geq 0\end{matrix}\right.$
$\to \left\{\begin{matrix}3x > 1\\ 4x^{2} \leq 1\end{matrix}\right.$
$\to \left\{\begin{matrix}x > \dfrac{1}{3}\\ -\dfrac{1}{2} \leq x \leq \dfrac{1}{2}\end{matrix}\right.$
$\to \dfrac{1}{3} < x \leq \dfrac{1}{2}$
+) $\left\{\begin{matrix}1 - 3x > 0\\ 1 - 4x^{2} > \left ( 1 - 3x \right )^{2}\end{matrix}\right.$
$\to \left\{\begin{matrix}3x < 1\\1 - 4x^{2} > 1 - 6x + 9x^{2}\end{matrix}\right.$
$\to \left\{\begin{matrix}x < \dfrac{1}{3}\\13x^{2} - 6x < 0\end{matrix}\right.$
$\to \left\{\begin{matrix}x < \dfrac{1}{3}\\0 < x < \dfrac{6}{13}\end{matrix}\right.$
$\to 0 < x < \dfrac{1}{3}$
$\to x \in \varnothing$
Kết hợp hai TH ta được $\to -\dfrac{1}{2} \leq x < 0$
ĐS: $-\dfrac{1}{2} \leq x < 0$
