Explicităm modulele:
[tex]\sqrt{6} - 2 = \sqrt{6} - \sqrt{4} > 0 \implies |\sqrt{6} - 2| = \sqrt{6} - 2[/tex]
[tex]\sqrt{6} - 3 = \sqrt{6} - \sqrt{9} < 0 \implies |\sqrt{6} - 3| = -(\sqrt{6} - 3) = 3 - \sqrt{6}[/tex]
[tex]\sqrt{6} - 4 = \sqrt{6} - \sqrt{16} < 0 \implies |\sqrt{6} - 4| = -(\sqrt{6} - 4) = 4 - \sqrt{6}[/tex]
[tex]\sqrt{6} + 1 = \sqrt{6} + \sqrt{1} > 0 \implies |\sqrt{6} + 1| = \sqrt{6} + 1[/tex]
Atunci, numărul a devine:
[tex]a = \sqrt{6} - 2 + 3 - \sqrt{6} + 4 - \sqrt{6} + \sqrt{6} + 1 = 6[/tex]
La numărul b avem:
[tex]b = |2\sqrt{3} - 3| + |\sqrt{3} - 5| + |2 - \sqrt{3}|[/tex]
Explicităm modulele:
[tex]2\sqrt{3} - 3 = \sqrt{12} - \sqrt{3} > 0 \implies |2\sqrt{3} - 3| = 2\sqrt{3} - 3[/tex]
[tex]\sqrt{3} - 5 = \sqrt{3} - \sqrt{25} < 0 \implies |\sqrt{3} - 5| = 5 - \sqrt{3}[/tex]
[tex]2 - \sqrt{3} = \sqrt{4} - \sqrt{3} > 0 \implies |2 - \sqrt{3}| = 2 - \sqrt{3}[/tex]
Atunci, numărul b devine:
[tex]b = 2\sqrt{3} - 3 + 5 - |\sqrt{3} + 2 - \sqrt{3} = 4[/tex]
⇒ a > b
