20. \( \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} \)

To simplify the expression: \[ \frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} \] we can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{2} + \sqrt{3}\) is \(\sqrt{2} - \sqrt{3}\). **Step 1: Multiply by the Conjugate** \[ \frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{\sqrt{6}(\sqrt{2} - \sqrt{3})}{(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})} \] **Step 2: Simplify the Denominator** \[ (\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1 \] **Step 3: Distribute in the Numerator** \[ \sqrt{6}(\sqrt{2} - \sqrt{3}) = \sqrt{6} \cdot \sqrt{2} - \sqrt{6} \cdot \sqrt{3} = \sqrt{12} - \sqrt{18} \] Simplify the square roots: \[ \sqrt{12} = 2\sqrt{3} \quad \text{and} \quad \sqrt{18} = 3\sqrt{2} \] So, \[ \sqrt{12} - \sqrt{18} = 2\sqrt{3} - 3\sqrt{2} \] **Step 4: Combine Everything** \[ \frac{2\sqrt{3} - 3\sqrt{2}}{-1} = -2\sqrt{3} + 3\sqrt{2} = 3\sqrt{2} - 2\sqrt{3} \] **Final Answer:** \[ \boxed{3\sqrt{2} - 2\sqrt{3}} \]