Explicație pas cu pas:
[tex]2 - 3 \sqrt{3} = \sqrt{4} - \sqrt{27} < 0 \implies |2 - 3 \sqrt{3}| = 3 \sqrt{3} - 2 \\ [/tex]
[tex]3 - 2 \sqrt{3} = \sqrt{9} - \sqrt{12} < 0 \implies |3 - 2 \sqrt{3}| = 2 \sqrt{3} - 3 \\ [/tex]
[tex]\sqrt{2} - \sqrt{3} < 0 \implies |\sqrt{2} - \sqrt{3}| = \sqrt{3} - \sqrt{2} \\ [/tex]
[tex]1 - \sqrt{2} = \sqrt{1} - \sqrt{2} < 0 \implies |1 - \sqrt{2}| = \sqrt{2} - 1 \\ [/tex]
a)
atunci:
[tex]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[/tex]
și
[tex]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[/tex]
b)
[tex]m_{g} = \sqrt{ab} = \sqrt{( \sqrt{3} + 1)( \sqrt{3} - 1)} = \\ = \sqrt{3 - 1} = \bf \sqrt{2}[/tex]
