Question

rationalise the denominator of the following
$\frac{1 + \sqrt{6} }{7 + 3 \sqrt{2} }$
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Answer 1

Answer:

It is the answer thanks

Answer 2

Step-by-step explanation:

$\frac{1 + \sqrt{6} }{7 + 3 \sqrt{2} } \\ \\ = \frac{1 + \sqrt{6} }{7 + 3 \sqrt{2} } \times \frac{7 - 3 \sqrt{2} }{7 - 3 \sqrt{2} } \\ \\ = \frac{(1 + \sqrt{6})(7 - 3 \sqrt{2} )}{ {(7)}^{2} - {(3 \sqrt{2}) }^{2} } \\ \\ = \frac{7 - 3 \sqrt{2} + 7 \sqrt{6} - 6 \sqrt{3} }{49 - 18} \\ \\ = \frac{7 - 3 \sqrt{2} + 7 \sqrt{6} - 6 \sqrt{3} }{31}$

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