1.6 Rationalising Denominators - 章节练习

答案：
a) $$\frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$
b) $$\frac{3}{2 + \sqrt{3}} = \frac{3(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{3(2 - \sqrt{3})}{4 - 3} = \frac{3(2 - \sqrt{3})}{1} = 6 - 3\sqrt{3}$$
c) $$\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} - \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{3 - 2\sqrt{6} + 2}{3 - 2} = \frac{5 - 2\sqrt{6}}{1} = 5 - 2\sqrt{6}$$

答案：
$$\frac{1}{\sqrt{2} + 1} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1$$
$$\frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1$$
相加：$$(\sqrt{2} - 1) + (\sqrt{2} + 1) = 2\sqrt{2}$$

答案：
先展开分母：$$(3 - \sqrt{2})^2 = 9 - 6\sqrt{2} + 2 = 11 - 6\sqrt{2}$$
有理化：$$\frac{5}{11 - 6\sqrt{2}} = \frac{5(11 + 6\sqrt{2})}{(11 - 6\sqrt{2})(11 + 6\sqrt{2})} = \frac{5(11 + 6\sqrt{2})}{121 - 72} = \frac{5(11 + 6\sqrt{2})}{49} = \frac{55 + 30\sqrt{2}}{49} = \frac{55}{49} + \frac{30\sqrt{2}}{49} = \frac{55 \div 1}{49 \div 1} + \frac{30\sqrt{2}}{49} = \frac{55}{49} + \frac{30\sqrt{2}}{49}$$
约分：$$\frac{55}{49} = \frac{55 \div 1}{49 \div 1} = \frac{55}{49}$$, 但不能再约分，所以最终结果为$$\frac{55 + 30\sqrt{2}}{49}$$

答案：
先有理化x：$$x = \frac{1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{\sqrt{3} + 1}{3 - 1} = \frac{\sqrt{3} + 1}{2}$$
代入计算：
$$x^2 = \left( \frac{\sqrt{3} + 1}{2} \right)^2 = \frac{3 + 2\sqrt{3} + 1}{4} = \frac{4 + 2\sqrt{3}}{4} = \frac{2 + \sqrt{3}}{2}$$
$$2x = 2 \times \frac{\sqrt{3} + 1}{2} = \sqrt{3} + 1$$
$$x^2 - 2x + 2 = \frac{2 + \sqrt{3}}{2} - (\sqrt{3} + 1) + 2 = \frac{2 + \sqrt{3}}{2} - \sqrt{3} - 1 + 2$$
$$= \frac{2 + \sqrt{3}}{2} + 1 - \sqrt{3} = \frac{2 + \sqrt{3} + 2 - 2\sqrt{3}}{2} = \frac{4 - \sqrt{3}}{2} = 2 - \frac{\sqrt{3}}{2}$$