1. **State the problem:** Simplify the expression $$\sqrt{3 + \sqrt{3}} - \sqrt{|1|} - \sqrt{1}$$. 2. **Recall important rules:** - The square root of an absolute value is the absolute value of the square root: $$\sqrt{|a|} = |\sqrt{a}|$$. - $$\sqrt{1} = 1$$. 3. **Simplify each term:** - $$\sqrt{|1|} = \sqrt{1} = 1$$. - $$\sqrt{1} = 1$$. 4. **Focus on simplifying $$\sqrt{3 + \sqrt{3}}$$:** We try to express it as $$\sqrt{a} + \sqrt{b}$$ for some $$a$$ and $$b$$. 5. **Set up the equation:** $$\sqrt{3 + \sqrt{3}} = \sqrt{a} + \sqrt{b}$$ Square both sides: $$3 + \sqrt{3} = a + b + 2\sqrt{ab}$$ 6. **Match terms:** - The rational part: $$a + b = 3$$ - The irrational part: $$2\sqrt{ab} = \sqrt{3}$$ 7. **Solve for $$a$$ and $$b$$:** From the irrational part: $$2\sqrt{ab} = \sqrt{3} \implies \sqrt{ab} = \frac{\sqrt{3}}{2} \implies ab = \frac{3}{4}$$ From the rational part: $$a + b = 3$$ 8. **Solve the system:** Let $$a$$ and $$b$$ be roots of $$x^2 - 3x + \frac{3}{4} = 0$$. Discriminant: $$\Delta = 9 - 3 = 6$$ Roots: $$x = \frac{3 \pm \sqrt{6}}{2}$$ 9. **Therefore:** $$\sqrt{3 + \sqrt{3}} = \sqrt{\frac{3 + \sqrt{6}}{2}} + \sqrt{\frac{3 - \sqrt{6}}{2}}$$ 10. **Substitute back and simplify the original expression:** $$\sqrt{3 + \sqrt{3}} - \sqrt{|1|} - \sqrt{1} = \left(\sqrt{\frac{3 + \sqrt{6}}{2}} + \sqrt{\frac{3 - \sqrt{6}}{2}}\right) - 1 - 1 = \sqrt{\frac{3 + \sqrt{6}}{2}} + \sqrt{\frac{3 - \sqrt{6}}{2}} - 2$$ **Final answer:** $$\boxed{\sqrt{\frac{3 + \sqrt{6}}{2}} + \sqrt{\frac{3 - \sqrt{6}}{2}} - 2}$$