\[\boxed{\text{439\ (439).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[Преобразуем\ все\ данные\ \]
\[числа.\]
\[15\sqrt{3} - 4\sqrt{2};\]
\[6 - \sqrt{12} = 6 - \sqrt{4} \cdot \sqrt{3} =\]
\[= 6 - 2\sqrt{3};\]
\[80 - 5\sqrt{3} = \sqrt{16 \cdot 5} - 5\sqrt{3} =\]
\[= 4\sqrt{5} - 5\sqrt{3};\]
\[75 - 4\sqrt{5} = \sqrt{25 \cdot 3} - 4\sqrt{5} =\]
\[= 5\sqrt{3} - 4\sqrt{5};\]
\[\frac{1}{2\sqrt{3} - 6} =\]
\[= \frac{2\sqrt{3} + 6}{\left( 2\sqrt{3} - 6 \right)\left( 2\sqrt{3} + 6 \right)} =\]
\[= \frac{2\sqrt{3} + 6}{4 \cdot 3 - 36} = \frac{2 \cdot \left( \sqrt{3} + 3 \right)}{- 24} =\]
\[= - \frac{\sqrt{3} + 3}{12};\]
\[\frac{1}{\sqrt{675} - \sqrt{32}} =\]
\[= \frac{\sqrt{675} + \sqrt{32}}{\left( \sqrt{675} - \sqrt{32} \right)\left( \sqrt{675} + \sqrt{32} \right)} =\]
\[= \frac{\sqrt{225 \cdot 3} + \sqrt{16 \cdot 2}}{675 - 32} =\]
\[= \frac{15\sqrt{3} + 4\sqrt{2}}{643}.\]
\[Числа\ \sqrt{80} - 5\sqrt{3}\ и\ \sqrt{75} - 4\sqrt{5} -\]
\[противоположные;\]
\[так\ как\ сумма\ чисел\ 4\sqrt{5} - 5\sqrt{3}\ \]
\[и\ 5\sqrt{3} - 4\sqrt{5}\ равна\ 0:\]
\[4\sqrt{5} - 5\sqrt{3} + 5\sqrt{3} - 4\sqrt{5} = 0.\]
\[\left( 15\sqrt{3} - 4\sqrt{2} \right) \cdot \frac{1}{\sqrt{675} - \sqrt{32}} =\]
\[= \frac{15\sqrt{3} - 4\sqrt{2}}{\sqrt{225 \cdot 3} - \sqrt{16 \cdot 2}} =\]
\[= \frac{15\sqrt{3} - 4\sqrt{2}}{15\sqrt{3} - 4\sqrt{2}} = 1;\]
\[Числа\ 15\sqrt{3\ } - 4\sqrt{2}\ и\ \]
\[\frac{1}{\sqrt{675} - \sqrt{32}} - взаимообратные,\ \]
\[так\ как\ числа\]
\[15\sqrt{3} - 4\sqrt{2}\ \ и\ \ \frac{15\sqrt{3} + 4\sqrt{2}}{643}\ \]
\[содержат\ одинаковые\ числа\ \]
\[15\sqrt{3}\ и\ 4\sqrt{2}:\]
\[\left( 15\sqrt{3} - 4\sqrt{2}\ \right) \cdot \frac{15\sqrt{3} + 4\sqrt{2}}{643} =\]
\[= \frac{\left( 15\sqrt{3} - 4\sqrt{2} \right)\left( 15\sqrt{3} + 4\sqrt{2} \right)}{643} =\]
\[= \frac{225 \cdot 3 - 16 \cdot 2}{643} = \frac{675 - 32}{643} =\]
\[= \frac{643}{643} = 1 - их\ произведение\ \]
\[равно\ 1.\ \]