Question

7<rad2+rad6+rad12<7.5 ​

Answer 1

Răspuns:

$7 < \sqrt{2} + \sqrt{6} + \sqrt{12} < 7.5 \\ \sqrt{2} + \sqrt{6} + \sqrt{12} = \sqrt{2} + \sqrt{6} + \sqrt{2 {}^{2} \times 3 } = \sqrt{2} + \sqrt{6} + 2 \sqrt{3} = 7.3278 \\ = > 7 < 7.3 < 7.5 \: (A)$

Answer 2

Răspuns:

Explicație pas cu pas: