Question

Simplify each of the following by rationalizing the denominator:
i) $\frac{6-4\sqrt{2}}{6+4\sqrt{2}}$ ii) $\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$
iii) $\frac{1}{3\sqrt{2}-2\sqrt{3}}$

Answer 1

$1)\frac{6-4\sqrt{2}}{6+4\sqrt{2}}=\frac{6-4\sqrt{2}}{6+4\sqrt{2}}\times \frac{6-4\sqrt{2}}{6-4\sqrt{2}}=\frac{(6-4\sqrt{2}{)}^2}{(6{)}^2-(4\sqrt{2}{)}^2}=\frac{36+32-48\sqrt{2}}{4}=17-12\sqrt{2}$

$2)\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}=\frac{(\sqrt{7}-\sqrt{5}{)}^2}{(\sqrt{7}{)}^2-(\sqrt{5}{)}^2}=\frac{7+5-2\sqrt{35}}{2}=6-\sqrt{15}$

$3)\frac{1}{3\sqrt{2}-2\sqrt{3}}=\frac{1}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}=\frac{3\sqrt{2}+2\sqrt{3}}{(3\sqrt{2}{)}^2-(2\sqrt{3}{)}^2}=\frac{3\sqrt{2}+2\sqrt{3}}{6}$