1. **Problem:** Simplify and analyze the expression $$\frac{1}{\sqrt{3 - \frac{x^2}{4}}}$$. 2. **Formula and rules:** The expression involves a square root in the denominator. Recall that $$\sqrt{a} = a^{1/2}$$ and the domain requires the radicand to be non-negative: $$3 - \frac{x^2}{4} \geq 0$$. 3. **Domain:** Solve $$3 - \frac{x^2}{4} \geq 0$$. Multiply both sides by 4: $$12 - x^2 \geq 0$$ Rearranged: $$x^2 \leq 12$$ So, $$-\sqrt{12} \leq x \leq \sqrt{12}$$ 4. **Simplify the expression:** Rewrite the denominator: $$\sqrt{3 - \frac{x^2}{4}} = \sqrt{\frac{12 - x^2}{4}} = \frac{\sqrt{12 - x^2}}{2}$$ Therefore, $$\frac{1}{\sqrt{3 - \frac{x^2}{4}}} = \frac{1}{\frac{\sqrt{12 - x^2}}{2}} = \frac{2}{\sqrt{12 - x^2}}$$ 5. **Final answer:** $$\boxed{\frac{2}{\sqrt{12 - x^2}}}$$ with domain $$-\sqrt{12} \leq x \leq \sqrt{12}$$. This expression is defined and real-valued only within this domain.