Question

rationalise the denominator ​

Answer 1

Answer:

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Step-by-step explanation:

$\frac{5 - \sqrt{27} }{ \sqrt{ 3} + \sqrt{5} } \\ \\ so \: conjugate \: of \: \sqrt{3} + \sqrt{5} \: \: is \: \: \sqrt{3} - \sqrt{5} \\ so \: rationalising \: denominator \\ we \: get \\ \\ = \frac{5 - \sqrt{27} }{ \sqrt{3} + \sqrt{5} } \: \times \frac{ \sqrt{3} - \sqrt{5} }{ \sqrt{ 3} - \sqrt{5} } \\ \\ = \frac{5 - 3 \sqrt{3} }{ \sqrt{ 3} + \sqrt{5} } \: \times \frac{ \sqrt{3} - \sqrt{5} }{ \sqrt{3} - \sqrt{5} } \\ \\ = \frac{(5 - 3 \sqrt{3} )( \sqrt{3} - \sqrt{5} )}{( \sqrt{3}) {}^{2} \: - ( \sqrt{5} ) {}^{2} } \\ \\ = \frac{5 \sqrt{3} - (3 \sqrt{3} \times \sqrt{3} ) - 3 \sqrt{3}.( - \sqrt{5} ) - 5 \sqrt{5} }{(3 - 5)} \\ \\ = \frac{5 \sqrt{ 3} - (3 \times 3) + 3 \sqrt{15} - 5 \sqrt{5} }{( - 2)} \\ \\ = \frac{5 \sqrt{3} - 9 + 3 \sqrt{15} - 5 \sqrt{5} }{( - 2)} \\ \\ = \frac{ - (5 \sqrt{3} + 3 \sqrt{15} - 5 \sqrt{5} + 9) }{2}$