Question

calculati |4-2radical6|+|3radical6-3|-|1-radical6|​

Answer 1

Explicație pas cu pas:

$4 - 2 \sqrt{6} = \sqrt{16} - \sqrt{24} < 0 \\ 3 \sqrt{6} - 3 = \sqrt{54} - \sqrt{9} > 0 \\ 1 - \sqrt{6} = \sqrt{1} - \sqrt{6} < 0$

$|4 - 2 \sqrt{6}| + |3 \sqrt{6} - 3| - |1 - \sqrt{6}| = \\ = 2 \sqrt{6} - 4 + 3 \sqrt{6} - 3 - ( \sqrt{6} - 1) \\ = 5 \sqrt{6} - 7 - \sqrt{6} + 1 = \bf 4 \sqrt{6} - 6$

Answer 2

$\sqrt4 < \sqrt6 \Rightarrow 2 < \sqrt6\bigg|_{\cdot2} \Rightarrow 4 < 2\sqrt6 \Rightarrow 4-2\sqrt6 < 0 \Rightarrow |4-2\sqrt6|=-4+3\sqrt6 \\ \\ |4-2\sqrt6|+|3\sqrt6-3|-|1-\sqrt6|=-4+2\sqrt6+3\sqrt6-3+1-\sqrt6=4\sqrt6-6$