Question

Ex 3 . Dau coroana…….

Answer 1

Explicație pas cu pas:

$2 - 3 \sqrt{3} = \sqrt{4} - \sqrt{27} < 0 \implies |2 - 3 \sqrt{3}| = 3 \sqrt{3} - 2$

$3 - 2 \sqrt{3} = \sqrt{9} - \sqrt{12} < 0 \implies |3 - 2 \sqrt{3}| = 2 \sqrt{3} - 3$

$\sqrt{2} - \sqrt{3} < 0 \implies |\sqrt{2} - \sqrt{3}| = \sqrt{3} - \sqrt{2}$

$1 - \sqrt{2} = \sqrt{1} - \sqrt{2} < 0 \implies |1 - \sqrt{2}| = \sqrt{2} - 1$

a)

atunci:

$a = \sqrt{ {(2 - 3 \sqrt{3} )}^{2} } - |3 - 2 \sqrt{3} | = |2 - 3 \sqrt{3}| - (2 \sqrt{3} - 3) = \\ = 3 \sqrt{3} - 2 - 2 \sqrt{3} + 3 = \bf \sqrt{3} + 1$

și

$b = | \sqrt{2} - \sqrt{3} | + \sqrt{ {(1 - \sqrt{2} )}^{2} } = \sqrt{3} - \sqrt{2} + |1 - \sqrt{2} | = \\ = \sqrt{3} - \sqrt{2} + \sqrt{2} - 1 = \bf \sqrt{3} - 1$

b)

$m_{g} = \sqrt{ab} = \sqrt{( \sqrt{3} + 1)( \sqrt{3} - 1)} = \\ = \sqrt{3 - 1} = \bf \sqrt{2}$