Question

Aratati ca:
$\sqrt{18} + \sqrt{72} - \sqrt{162} = 0$
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Answer 1

3 rad2 + 6 rad2 - 9 rad2 = 9 rad2 - 9 rad2 = 0

Answer 2

Răspuns:

$\sqrt{18} + \sqrt{72} - \sqrt{162} = 0 \\ \sqrt{9 \times 2} + \sqrt{9 \times 8} - \sqrt{81 \times 2} = 0 \\ 3 \sqrt{2} + 3 \sqrt{8} - 9 \sqrt{2} = 0 \\ 3 \sqrt{2} + 3 \times \sqrt{4 \times 2} - 9 \sqrt{2} = 0 \\ 3 \sqrt{2} + 3 \times 2 \sqrt{2} - 9 \sqrt{2} = 0 \\ 3 \sqrt{2} + 6 \sqrt{2} - 3 \sqrt{2} = 0 \\ 9 \sqrt{2} - 3 \sqrt{2} = 6 \sqrt{2} \\ = > 3 \sqrt{2} - 6 \sqrt{2} + 3 \sqrt{2} = 0 \\ - 3 \sqrt{2} + 3 \sqrt{2} = 0 \sqrt{2} = 0$