Rationalise the denominator in each of the following:
$\frac{18}{3 \sqrt{2}-2 \sqrt{3}}$
$\frac{18}{3 \sqrt{2}-2 \sqrt{3}}$
(N/A) To rationalise the denominator,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(3 \sqrt{2} + 2 \sqrt{3})$.
$\frac{18}{3 \sqrt{2}-2 \sqrt{3}} \times \frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2})^2 - (2 \sqrt{3})^2}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{(9 \times 2) - (4 \times 3)}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{18 - 12}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{6}$
$= 3(3 \sqrt{2}+2 \sqrt{3})$
$\frac{18}{3 \sqrt{2}-2 \sqrt{3}} \times \frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2})^2 - (2 \sqrt{3})^2}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{(9 \times 2) - (4 \times 3)}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{18 - 12}$
$= \frac{18(3 \sqrt{2}+2 \sqrt{3})}{6}$
$= 3(3 \sqrt{2}+2 \sqrt{3})$
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