Question

rationalise the denominator 16/4√3-√5 b) 2/2 √2-√7​

Answer 1

Solutions :-

1.

$\frac{16}{4 \sqrt{3} - \sqrt{5} } \\ \\ \frac{16}{ 4\sqrt{3} - \sqrt{5} } \times \frac{(4 \sqrt{3} + \sqrt{5} )}{(4 \sqrt{3} + \sqrt{5} )} \\ \\ \frac{ {16(4 \sqrt{3} + \sqrt{5} )} }{ ({4 \sqrt{3}) }^{2} - {( \sqrt{5}) }^{2} } \\ \\ \frac{ {16(4 \sqrt{3} + \sqrt{5} )} }{48 - 5} = \frac{ {16(4 \sqrt{3} + \sqrt{5} )} }{43}$

2.

$\frac{2}{2 - \sqrt{2} - \sqrt{7} } = \: \: \: \: \: \: \: \: \: \: \: \: (taking \: 2 - \sqrt{2} \: as \: a \: and \: 7 \: as \: b \: and \: making \: the \: {a}^{2} - {b}^{2} ) \\ \\ \frac{2}{2 - \sqrt{2} - \sqrt{7} } \times \frac{2 - \sqrt{2} + 7}{2 - \sqrt{2} + 7} \\ \\ \frac{2(2 - \sqrt{2} + 7)}{( {2 - \sqrt{2}) }^{2} - {(7)}^{2} } \\ \\ \frac{2(2 - \sqrt{2 } + 7) }{6 - 4 \sqrt{2} - 49}$

Rationalise it again. Sorry i am unable to do as this app is crashing again and again.

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