Question

Rationalise the denominator

Answer 1

rationalise the three terms individually

Answer 2

ϲοиѕι∂єя єգ (1) , єգ (2) αи∂ єգ(3)

$\frac{ \sqrt[2]{6} }{ \sqrt{2} + \sqrt{3} } + \frac{ \sqrt[6]{2} }{ \sqrt{6 + \sqrt{3} } } - \frac{ \sqrt[8]{3} }{ \sqrt{6} + \sqrt{2} }$

єգ ( 1 )

$\frac{ \sqrt[2]{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2 - \sqrt{3} } }{ \sqrt{2 - \sqrt{3} } }$

$\frac{2 \sqrt{6} {( \sqrt{2 - \sqrt{3}) } } }{ \sqrt{ {2}^{2} } - \sqrt{ {3}^{2} } } = \frac{ \sqrt[2]{6}( \sqrt{2} - \sqrt{3}) }{2 - 3}$

$\frac{ \sqrt{6}( \sqrt{2} - \sqrt{3} )}{3} = \frac{ \sqrt{12} - \sqrt{18} }{3}$

ѕιмιℓαяℓγ.........

єգ(2)

$\frac{ \sqrt{12} - \sqrt{6} }{3}$

(єգ 3)

$\frac{ \sqrt[8]{18} - \sqrt[8]{24} }{4}$

ρυττιиg єg (1),(2) αи∂ (3) τοgєτнєя

$\frac{ \sqrt{12} - \sqrt{18} }{3} + \frac{ \sqrt{12} - \sqrt{6} }{3} - \frac{ \sqrt[8]{18} - \sqrt[8]{24} }{4}$

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