Question

18 supra radical din6 minus 5 supra radical din 6 minus 1 minus radical din 24​

Answer 1

${}^{ {}^{ \sqrt{6} )} } \frac{18}{ \sqrt{6} } - {}^{ {}^{ \sqrt{6}) } } \frac{5}{ \sqrt{6} } - \frac{1}{ \sqrt{24} } \\ \\ = {}^{ {}^{2)} } \frac{18 \sqrt{6} }{6} - {}^{ {}^{2)} } \frac{5 \sqrt{6} }{6} - {}^{ {}^{ \sqrt{6} )} } \frac{1}{2 \sqrt{6} } \\ \\ = \frac{36 \sqrt{6} }{12} - \frac{10 \sqrt{6} }{12} - \frac{ \sqrt{6} }{12} \\ \\ = \frac{25 \sqrt{6} }{12}$

Answer 2

Explicație pas cu pas:

$\dfrac{18}{ \sqrt{6} } - \dfrac{5}{ \sqrt{6} - 1} - \sqrt{24} =$

$= \dfrac{18 \sqrt{6} }{ {( \sqrt{6} )}^{2} } - \dfrac{5( \sqrt{6} + 1)}{( \sqrt{6} - 1)( \sqrt{6} + 1)} - \sqrt{ {2}^{2} \cdot 6}$

$= \dfrac{18 \sqrt{6} }{6} - \dfrac{5( \sqrt{6} + 1)}{6 - 1} - 2\sqrt{6}$

$= 3 \sqrt{6} - \dfrac{5( \sqrt{6} + 1)}{5} - 2\sqrt{6}$

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

$= 3 \sqrt{6} - \sqrt{6} - 1 - 2\sqrt{6}$

$= 3 \sqrt{6} - 3 \sqrt{6} - 1 = \bf - 1$