Question

30 БАЛОВ!! Довести тотожність
$\ \textless \ br /\ \textgreater \ ( \frac{3}{3 + \sqrt{6} } + \frac{2}{3 - \sqrt{6} } ) \times \frac{3}{15 - \sqrt{6} } = 1$
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Answer 1

$\displaystyle \bigg( \frac{3}{3+\sqrt{6}}+\frac{2}{3-\sqrt{6}} \bigg)\cdot \frac{3}{15-\sqrt{6}}=\bigg(\frac{3(3-\sqrt{6})}{(3+\sqrt{6})(3-\sqrt{6})}+\frac{2(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\bigg)\cdot\\\\\\\cdot\frac{3}{15-\sqrt{6}}=\bigg(3-\sqrt{6}+\frac{2(3+\sqrt{6})}{3}\bigg)\cdot \frac{3}{15-\sqrt{6}}=\frac{9-3\sqrt{6}+6+2\sqrt{6}}{3}\cdot \\\\\\\cdot\frac{3}{15-\sqrt{6}}=\frac{15-\sqrt{6}}{3}\cdot \frac{3}{15-\sqrt{6}}=1$