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July 2022 1 17 Report

Освободитесь от иррациональности в знаменателе дроби:
[tex]\frac{6}{\sqrt{10}+\sqrt{6}+5-\sqrt{15}} [/tex]

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$\frac{6}{\sqrt{10}+\sqrt{6}+5-\sqrt{15}} =
\frac{6}{(\sqrt{10}+\sqrt{6}) + (5-\sqrt{15})} = \\ \\ \\ =
\frac{6}{(\sqrt{10}+\sqrt{6}) + (5-\sqrt{15})} * \frac{(\sqrt{10}+\sqrt{6}) -
(5-\sqrt{15})}{(\sqrt{10}+\sqrt{6}) - (5-\sqrt{15})} = \\ \\ \\ = \frac{6((\sqrt{10}+\sqrt{6})
- (5-\sqrt{15}))}{(\sqrt{10}+\sqrt{6})^2 - (5-\sqrt{15})^2} = \\ \\ \\ =
\frac{6((\sqrt{10}+\sqrt{6}) - (5-\sqrt{15}))}{(16+2\sqrt{60}) -
(40-10\sqrt{15})} =$

$= \frac{6(\sqrt{10}+\sqrt{6} -5+\sqrt{15})}{14\sqrt{15} - 24} = \\ \\ \\ = \frac{6(\sqrt{10}+\sqrt{6} -5+\sqrt{15})}{14\sqrt{15} - 24} * \frac{{14 \sqrt{15} + 24}}{14 \sqrt{15} + 24}} = \\ \\ \\ = \frac{6(\sqrt{10}+\sqrt{6} -5+\sqrt{15})*({14 \sqrt{15} + 24})}{(14\sqrt{15})^2 - 24^2} =$

$= \frac{6(\sqrt{10}+\sqrt{6} -5+\sqrt{15})*({14 \sqrt{15} + 24})}{2364}$

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