423. Сократите дробь: a) \frac{x^2-2}{x+\sqrt{2}}; б) \frac{\sqrt{5}-a}{5-a^2}; в) \frac{\sqrt{x}-5}{25-x}; г) \frac{\sqrt{2}+2}{\sqrt{2}}; д) \frac{5+\sqrt{10}}{\sqrt{10}}; e) \frac{2\sqrt{3}-3}{5\sqrt{3}}

Давай сократим каждую дробь по порядку: a) \(\frac{x^2-2}{x+\sqrt{2}} = \frac{(x-\sqrt{2})(x+\sqrt{2})}{x+\sqrt{2}} = x - \sqrt{2}\) б) \(\frac{\sqrt{5}-a}{5-a^2} = \frac{\sqrt{5}-a}{(\sqrt{5}-a)(\sqrt{5}+a)} = \frac{1}{\sqrt{5}+a}\) в) \(\frac{\sqrt{x}-5}{25-x} = \frac{\sqrt{x}-5}{(5-\sqrt{x})(5+\sqrt{x})} = -\frac{1}{5+\sqrt{x}}\) г) \(\frac{\sqrt{2}+2}{\sqrt{2}} = \frac{\sqrt{2}(1+\sqrt{2})}{\sqrt{2}} = 1 + \sqrt{2}\) д) \(\frac{5+\sqrt{10}}{\sqrt{10}} = \frac{5}{\sqrt{10}} + 1 = \frac{5\sqrt{10}}{10} + 1 = \frac{\sqrt{10}}{2} + 1\) e) \(\frac{2\sqrt{3}-3}{5\sqrt{3}} = \frac{2\sqrt{3}}{5\sqrt{3}} - \frac{3}{5\sqrt{3}} = \frac{2}{5} - \frac{3\sqrt{3}}{5 \cdot 3} = \frac{2}{5} - \frac{\sqrt{3}}{5}\)

Ответ:

a) \(x - \sqrt{2}\)

б) \(\frac{1}{\sqrt{5}+a}\)

в) \(-\frac{1}{5+\sqrt{x}}\)

г) \(1 + \sqrt{2}\)

д) \(\frac{\sqrt{10}}{2} + 1\)

e) \(\frac{2}{5} - \frac{\sqrt{3}}{5}\)

Ты молодец! У тебя всё получится!