Question

$\frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-2\sqrt{3}}$ equals

• A. $0$
• B. $1$
• C. $-2$
• D. $4\sqrt{6}$

Answer 1

Answer: (D)

$4\sqrt{6}$

$\frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-2\sqrt{3}}$
$=\frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-3\sqrt{2}}-\frac{\sqrt{18}}{3-2\sqrt{3}}$
$=\frac{4\sqrt{3}(2+\sqrt{2})}{2-\sqrt{2}(2+\sqrt{2})}-\frac{30(4\sqrt{3}+3\sqrt{2})}{\left(\right(4\sqrt{3})-3\sqrt{2})(4\sqrt{3}+3\sqrt{2})}-\frac{\sqrt{18}(3+2\sqrt{3})}{(3-2\sqrt{3})(3+2\sqrt{3})}$
$=\frac{4\sqrt{3}(2+\sqrt{2})}{4-2}-\frac{30(4\sqrt{3}+3\sqrt{2})}{48-18}-\frac{\sqrt{18}(3+2\sqrt{3})}{9-12}$
$=\frac{4\sqrt{3}(2+\sqrt{2})}{2}-\frac{30(4\sqrt{3}+3\sqrt{2})}{30}+\frac{3\sqrt{2}(3+2\sqrt{3})}{3}$
$=2\sqrt{3}(2+\sqrt{2})-(4\sqrt{3}+3\sqrt{2})+\sqrt{2}(3+2\sqrt{3})$
$=4\sqrt{3}+2\sqrt{6}-4\sqrt{3}-3\sqrt{2}+3\sqrt{2}+2\sqrt{6}$
$=4\sqrt{6}$