Question: top-left: Rationalize the denominator of $\frac{\sqrt{k}}{\sqrt{k} + \sqrt{g}}$. Assume that all variables represent positive real numbers.
bottom-left: $\frac{\sqrt{k}}{\sqrt{k} + \sqrt{g}} = \square$
1. **State the problem:** Rationalize the denominator of the expression $$\frac{\sqrt{k}}{\sqrt{k} + \sqrt{g}}$$ where $k$ and $g$ are positive real numbers.
2. **Recall the formula:** To rationalize a denominator of the form $a + b$, multiply numerator and denominator by the conjugate $a - b$ to eliminate the square roots in the denominator.
3. **Apply the conjugate:** Multiply numerator and denominator by $$\sqrt{k} - \sqrt{g}$$:
$$\frac{\sqrt{k}}{\sqrt{k} + \sqrt{g}} \times \frac{\sqrt{k} - \sqrt{g}}{\sqrt{k} - \sqrt{g}} = \frac{\sqrt{k}(\sqrt{k} - \sqrt{g})}{(\sqrt{k} + \sqrt{g})(\sqrt{k} - \sqrt{g})}$$
4. **Simplify the numerator:**
$$\sqrt{k} \times \sqrt{k} - \sqrt{k} \times \sqrt{g} = k - \sqrt{kg}$$
5. **Simplify the denominator using difference of squares:**
$$(\sqrt{k})^2 - (\sqrt{g})^2 = k - g$$
6. **Write the rationalized expression:**
$$\frac{k - \sqrt{kg}}{k - g}$$
7. **Final answer:**
$$\boxed{\frac{k - \sqrt{kg}}{k - g}}$$