Questions

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

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a
$\sqrt{11}-\sqrt{5}+3\sqrt{2}-1$
b
$\sqrt{12}+\sqrt{5}+2\sqrt{2}-13$
c
$\sqrt{10}-\sqrt{5}+2\sqrt{2}-1$
d
$\sqrt{13}-\sqrt{3}+2\sqrt{3}-1$
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detailed solution

Correct option is C

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

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

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

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

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

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

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