Question
Simplify by rationalising the denominator: $\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$

Answer

$\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$
$=\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{16\times3}+\sqrt{9\times2}}$
$=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}$
$=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\times\frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}}$ $=\frac{7\sqrt{3}\times4\sqrt{3}-7\sqrt{3}\times3\sqrt{2}-5\sqrt{2}\times4\sqrt{3}+5\sqrt{2}\times3\sqrt{2}}{\big(4\sqrt{3}\big)^2-\big(3\sqrt{2}\big)^2}$
$=\frac{28\times3-21\sqrt{6}-20\sqrt{6}+15\times2}{16\times3-9\times2}$
$=\frac{84-41\sqrt{6}+30}{48-18}$
$=\frac{114-41\sqrt{6}}{30}$

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