Question

Evaluate: $\frac{\sqrt{2}+\sqrt{2}}{7+\sqrt{7}}÷\frac{\sqrt{5}+\sqrt{4}}{\sqrt{2}+3}$

Answer 1

$\frac{\sqrt{2}+\sqrt{2}}{7+\sqrt{2}}\times \frac{\sqrt{2}+3}{\sqrt{5}+\sqrt{4}}$

Rationalising the denominator, we get

$=\frac{2\sqrt{2}}{7+\sqrt{2}}\times \frac{7-\sqrt{2}}{7-\sqrt{2}}\times \frac{\sqrt{2}+3}{\sqrt{5}+2}\times \frac{\sqrt{5}-2}{\sqrt{5}-2}$

$=\frac{2\sqrt{2}(7-\sqrt{2})}{49-2}\times \frac{(\sqrt{2}+3)(\sqrt{5}-2)}{5-4}$

$=\frac{(14\sqrt{2}-4)}{47}\times (\sqrt{10}-2\sqrt{2}+3\sqrt{5}-6)$

$=\frac{1}{47}\times 2(7\sqrt{2}-2)(\sqrt{10}-2\sqrt{2}+3\sqrt{5}-6)$

$=\frac{2}{47}(7\sqrt{2}(\sqrt{10}-2\sqrt{2}+3\sqrt{5}-6)-2(\sqrt{10}-2\sqrt{2}+3\sqrt{5}-6))$

$=\frac{2}{47}(7\sqrt{20}-28+21\sqrt{10}-42\sqrt{2}-2\sqrt{10}+4\sqrt{2}-6\sqrt{5}+12)$

$=\frac{2}{47}(14\sqrt{5}-28+21\sqrt{10}-42\sqrt{2}-2\sqrt{10}+4\sqrt{2}-6\sqrt{5}+12)$

$=\frac{2}{47}(8\sqrt{5}+19\sqrt{10}-38\sqrt{2}-16)$

$=\frac{2}{47}[\sqrt{5}(8+19\sqrt{2})-2(19\sqrt{2}+8)]$

$=\frac{2}{47}(8+19\sqrt{2})(\sqrt{5}-2)$