$$63+2$$−$$5$$×$$3+2+53+2+5=6(3+2+5)(3+2)2$$−$$(5)2=6(3+2+5)9+2+62$$−$$5=6(3+2+5)6+62=6(3+2+5)6(1+2)=3+2+51+2$$×$$2$$−$$12$$−$$1=32$$−$$3+2$$−$$2+10$$−$$51=22+10$$−$$5$$−$$1$$

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$\frac{6}{3 + \sqrt{2} - \sqrt{5}} =$

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\sqrt{11} - \sqrt{5} + 3 \sqrt{2} - 1

\sqrt{12} + \sqrt{5} + 2 \sqrt{2} - 13

\sqrt{10} - \sqrt{5} + 2 \sqrt{2} - 1

\sqrt{13} - \sqrt{3} + 2 \sqrt{3} - 1

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detailed solution

Correct option is C

$\frac{6}{3 + \sqrt{2} - \sqrt{5}} \times \frac{3 + \sqrt{2} + \sqrt{5}}{3 + \sqrt{2} + \sqrt{5}}$

$= \frac{6 (3 + \sqrt{2} + \sqrt{5})}{{(3 + \sqrt{2})}^{2} - {(\sqrt{5})}^{2}} = \frac{6 (3 + \sqrt{2} + \sqrt{5})}{9 + 2 + 6 \sqrt{2} - 5}$

$= \frac{6 (3 + \sqrt{2} + \sqrt{5})}{6 + 6 \sqrt{2}} = \frac{6 (3 + \sqrt{2} + \sqrt{5})}{6 (1 + \sqrt{2})}$

$= \frac{3 + \sqrt{2} + \sqrt{5}}{1 + \sqrt{2}} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}$

$= \frac{3 \sqrt{2} - 3 + 2 - \sqrt{2} + \sqrt{10} - \sqrt{5}}{1}$

$= 2 \sqrt{2} + \sqrt{10} - \sqrt{5} - 1$

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63+2−5=

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$\frac{6}{3 + \sqrt{2} - \sqrt{5}} =$