Question

rationlized the denomineter 6 upon
2√3+√6​

Answer 1

Answer:

$\frac{6}{2 \sqrt{3} + \sqrt{6}} = 2 \sqrt{3} - \sqrt{6}$

Step-by-step explanation:

$\frac{6}{2 \sqrt{3} + \sqrt{6}} \\ = \frac{6}{2 \sqrt{3} + \sqrt{6} } \times \frac{2 \sqrt{3} - \sqrt{6} }{2 \sqrt{3} - \sqrt{6} } \\ = \frac{6(2 \sqrt{3} - \sqrt{6}) }{(2 \sqrt{3} + \sqrt{6} )(2 \sqrt{3} - \sqrt{6} ) } \\ = \frac{6(2 \sqrt{3} - \sqrt{6}) }{ {(2 \sqrt{3}) }^{2} - {( \sqrt{6}) }^{2} } \\ = \frac{6(2 \sqrt{3} - \sqrt{6}) }{12 - 6} \\ = \frac{6(2 \sqrt{3} - \sqrt{6})}{6} \\ = 2 \sqrt{3} - \sqrt{6}$