Question

radical (4radical din 2 - 6)la puterea 2+radical(3radical 2-4) la puterea 2​

Answer 1

Explicație pas cu pas:

$4 \sqrt{2} - 6 = \sqrt{32} - \sqrt{36} < 0 \implies |4 \sqrt{2} - 6| = 6 - 4 \sqrt{2}$

$3 \sqrt{2} - 4 = \sqrt{18} - \sqrt{16} > 0 \implies |3 \sqrt{2} - 4| = 3 \sqrt{2} - 4$

=>

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