Question

Buna ma ajutati cu o rezolvare

Answer 1

$a) \frac{6}{2 \sqrt{3} - \sqrt{6} } - \sqrt{12} + 1 \\ \frac{6(2 \sqrt{3} + \sqrt{6)} }{(2 \sqrt{3} - \sqrt{6} )(2 \sqrt{3} + \sqrt{6}) } - \sqrt{12} + 1 = \\ \frac{12 \sqrt{3} + 6 \sqrt{6} }{12 - 6} - \sqrt{12} + 1 = \\ \frac{12 \sqrt{3} + 6 \sqrt{6} }{6} - \frac{12 \sqrt{3} }{6} + \frac{6}{6} = \\ \frac{6 \sqrt{6} + 6}{6} = \frac{6( \sqrt{6} + 1) }{6} = \sqrt{6} + 1$

$b) \sqrt{ \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } } = \sqrt{ \frac{( \sqrt{5} - \sqrt{3})( \sqrt{5} - \sqrt{3} ) }{( \sqrt{5} + \sqrt{3} )( \sqrt{5} - \sqrt{3}) } } = \\ \sqrt{ \frac{5 - \sqrt{15} - \sqrt{15} + 3 }{5 - 3} } = \sqrt{ \frac{8 - 2 \sqrt{15} }{2} } = \\ \sqrt{ \frac{2(4 - \sqrt{15}) }{2} } = \sqrt{4 - \sqrt{15} }$