1. **Problem:** Simplify the expression $$\sqrt{2 - \sqrt{3}}(\sqrt{2} + \sqrt{3}) - 2(\sqrt{2} + \sqrt{3})$$. 2. **Step 1:** Let $$a = \sqrt{2 - \sqrt{3}}$$ and $$b = \sqrt{2} + \sqrt{3}$$. 3. **Step 2:** The expression becomes $$a \cdot b - 2b = b(a - 2)$$. 4. **Step 3:** Calculate $$a$$ by simplifying $$\sqrt{2 - \sqrt{3}}$$. Note that $$2 - \sqrt{3} = \left(\sqrt{3} - 1\right)^2$$ because: $$\left(\sqrt{3} - 1\right)^2 = (\sqrt{3})^2 - 2 \times \sqrt{3} \times 1 + 1^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}$$ which is not equal to $$2 - \sqrt{3}$$. So, try another approach. 5. **Step 4:** Use the conjugate to simplify the product $$(\sqrt{2 - \sqrt{3}})(\sqrt{2} + \sqrt{3})$$. Alternatively, first rationalize or express $$\sqrt{2 - \sqrt{3}}$$ in a simpler nested radical form: Assume $$\sqrt{2 - \sqrt{3}} = \sqrt{a} - \sqrt{b}$$ for some positive $$a,b$$. Square both sides: $$2 - \sqrt{3} = a + b - 2 \sqrt{ab}$$. Matching terms gives: $$a + b = 2$$ and $$2 \sqrt{ab} = \sqrt{3} \Rightarrow 4ab = 3 \Rightarrow ab = \frac{3}{4}$$. 6. **Step 5:** Solve for $$a, b$$: From $$a + b = 2$$ and $$ab = \frac{3}{4}$$, the quadratic equation is: $$x^2 - 2x + \frac{3}{4} = 0$$. Discriminant: $$\Delta = 4 - 3 = 1$$. Roots: $$x = \frac{2 \pm 1}{2}$$ gives $$x = \frac{3}{2}$$ or $$x = \frac{1}{2}$$. So $$a = \frac{3}{2}$$ and $$b = \frac{1}{2}$$ (order doesn't matter). 7. **Step 6:** Then: $$\sqrt{2 - \sqrt{3}} = \sqrt{\frac{3}{2}} - \sqrt{\frac{1}{2}} = \frac{\sqrt{3} - 1}{\sqrt{2}}$$. 8. **Step 7:** Substitute back into the original expression: $$\left( \frac{\sqrt{3} - 1}{\sqrt{2}} \right)(\sqrt{2} + \sqrt{3}) - 2(\sqrt{2} + \sqrt{3})$$. 9. **Step 8:** Multiply: $$\frac{\sqrt{3} - 1}{\sqrt{2}} \times (\sqrt{2} + \sqrt{3}) = \frac{(\sqrt{3} - 1)(\sqrt{2} + \sqrt{3})}{\sqrt{2}}$$. Calculate numerator: $$(\sqrt{3} \times \sqrt{2}) + (\sqrt{3} \times \sqrt{3}) - (1 \times \sqrt{2}) - (1 \times \sqrt{3}) = \sqrt{6} + 3 - \sqrt{2} - \sqrt{3}$$. So expression is: $$\frac{\sqrt{6} + 3 - \sqrt{2} - \sqrt{3}}{\sqrt{2}} - 2(\sqrt{2} + \sqrt{3})$$. 10. **Step 9:** Write $$2(\sqrt{2} + \sqrt{3})$$ as: $$2\sqrt{2} + 2\sqrt{3}$$. 11. **Step 10:** The full expression is: $$\frac{\sqrt{6} + 3 - \sqrt{2} - \sqrt{3}}{\sqrt{2}} - 2\sqrt{2} - 2\sqrt{3}$$. 12. **Step 11:** Split the fraction: $$\frac{\sqrt{6}}{\sqrt{2}} + \frac{3}{\sqrt{2}} - \frac{\sqrt{2}}{\sqrt{2}} - \frac{\sqrt{3}}{\sqrt{2}} - 2\sqrt{2} - 2\sqrt{3}$$ Simplify terms: - $$\frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$$ - $$\frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$$ (rationalized) - $$\frac{\sqrt{2}}{\sqrt{2}} = 1$$ - $$\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2}$$ (rationalized) So it becomes: $$\sqrt{3} + \frac{3\sqrt{2}}{2} - 1 - \frac{\sqrt{6}}{2} - 2\sqrt{2} - 2\sqrt{3}$$ 13. **Step 12:** Group like terms: $$(\sqrt{3} - 2\sqrt{3}) + \left(\frac{3\sqrt{2}}{2} - 2\sqrt{2}\right) - 1 - \frac{\sqrt{6}}{2}$$ Which is: $$-\sqrt{3} + \left(\frac{3\sqrt{2}}{2} - \frac{4\sqrt{2}}{2} \right) - 1 - \frac{\sqrt{6}}{2} = -\sqrt{3} - \frac{\sqrt{2}}{2} - 1 - \frac{\sqrt{6}}{2}$$ 14. **Final Answer:** $$-1 - \sqrt{3} - \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2}$$ This is the simplified form of the original expression.