Rationalise the denominator of the following expression:
$\frac{30}{5 \sqrt{3}-3 \sqrt{5}}$
$\frac{30}{5 \sqrt{3}-3 \sqrt{5}}$
(N/A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(5 \sqrt{3} + 3 \sqrt{5})$.
$\frac{30}{5 \sqrt{3}-3 \sqrt{5}} = \frac{30}{5 \sqrt{3}-3 \sqrt{5}} \times \frac{5 \sqrt{3}+3 \sqrt{5}}{5 \sqrt{3}+3 \sqrt{5}}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{(5 \sqrt{3})^{2}-(3 \sqrt{5})^{2}}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{75-45}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{30}$
$= 5 \sqrt{3}+3 \sqrt{5}$
$\frac{30}{5 \sqrt{3}-3 \sqrt{5}} = \frac{30}{5 \sqrt{3}-3 \sqrt{5}} \times \frac{5 \sqrt{3}+3 \sqrt{5}}{5 \sqrt{3}+3 \sqrt{5}}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{(5 \sqrt{3})^{2}-(3 \sqrt{5})^{2}}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{75-45}$
$= \frac{30(5 \sqrt{3}+3 \sqrt{5})}{30}$
$= 5 \sqrt{3}+3 \sqrt{5}$
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