1. **State the problem:** Simplify the expression $$\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}$$ step by step. 2. **Recall the formula:** To simplify a fraction with radicals in the denominator, multiply numerator and denominator by the conjugate of the denominator. The conjugate of $$\sqrt{12} + \sqrt{3}$$ is $$\sqrt{12} - \sqrt{3}$$. 3. **Multiply numerator and denominator by the conjugate:** $$\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}} \times \frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} - \sqrt{3}} = \frac{(\sqrt{12} - \sqrt{3})^2}{(\sqrt{12} + \sqrt{3})(\sqrt{12} - \sqrt{3})}$$ 4. **Expand numerator using the formula $ (a - b)^2 = a^2 - 2ab + b^2 $:** $$ (\sqrt{12})^2 - 2 \times \sqrt{12} \times \sqrt{3} + (\sqrt{3})^2 = 12 - 2\sqrt{36} + 3 $$ 5. **Simplify the terms:** $$ 12 - 2 \times 6 + 3 = 12 - 12 + 3 = 3 $$ 6. **Simplify denominator using difference of squares formula $ (a + b)(a - b) = a^2 - b^2 $:** $$ (\sqrt{12})^2 - (\sqrt{3})^2 = 12 - 3 = 9 $$ 7. **Write the fraction with simplified numerator and denominator:** $$ \frac{3}{9} $$ 8. **Simplify the fraction by dividing numerator and denominator by 3:** $$ \frac{\cancel{3}}{\cancel{9}} = \frac{1}{3} $$ **Final answer:** $$\frac{1}{3}$$