Question

simplify by rationalizing the denominator

Answer 1

Heya there !!!
Here is the answer you were looking for:

Identity used :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$
$\frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } + \frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } + \frac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \\ \\ = \frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } + \frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } \times \frac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} } + \frac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \times \frac{ \sqrt{6} - \sqrt{2} }{ \sqrt{6} - \sqrt{2} } \\ \\ = \frac{ \sqrt{6}( \sqrt{2} - \sqrt{3} )}{ {( \sqrt{2}) }^{2} - {( \sqrt{3}) }^{2} } + \frac{3 \sqrt{2}( \sqrt{6} + \sqrt{3} ) }{ {( \sqrt{6}) }^{2} - { \sqrt{3}) }^{2} } + \frac{4 \sqrt{3}( \sqrt{6} - \sqrt{2} ) }{ {( \sqrt{6}) }^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{ \sqrt{12} - \sqrt{18} }{2 - 3} + \frac{3 \sqrt{12} + 3 \sqrt{6} }{6 - 3} + \frac{4 \sqrt{18} - 4 \sqrt{6} }{6 - 2} \\ \\ = - ( \sqrt{2 \times 2 \times 3} - \sqrt{3 \times 3 \times 2} ) + \frac{3 \sqrt{2 \times 2 \times 3 } + 3 \sqrt{6} }{3} + \frac{4 \sqrt{3 \times 3 \times 2} - 4 \sqrt{6} }{4} \\ \\ = - \sqrt{ {2}^{2} \times 3 } + \sqrt{ {3}^{2} \times 2} + \sqrt{ {2}^{2} \times 3 } + \sqrt{6} + \sqrt{ {3}^{2} \times 2} - \sqrt{6} \\ \\ = - 2 \sqrt{3} + 3 \sqrt{2} + 2 \sqrt{3} + \sqrt{6} + 3 \sqrt{2} - \sqrt{6} \\ \\ = 0$

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@Mahak24

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Answer 2

Heya!!!!

Here's your answer =>=>

Refer to the attached file ^^^^

Plz ask me if you feel any problem.

Hope it will help you ☆▪☆

Thanks ^_^