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| = 3 - \sqrt{6}[/tex]
[tex]\sqrt{6} - 4 = \sqrt{6} - \sqrt{16} < 0 \implies |\sqrt{6} - 4| = 4 - \sqrt{6}[/tex]
[tex]\sqrt{6} + 1 > 0 \implies |\sqrt{6} + 1| = \sqrt{6} + 1[/tex]
Numărul a este:
[tex]a = \sqrt{6} - 2 + 3 - \sqrt{6} + 4 - \sqrt{6} + \sqrt{6} + 1 = (2\sqrt{6} - 2\sqrt{6}) + (8 - 2) = 6[/tex]
Numărul b este:
[tex]b = | 2\sqrt{3} - 3| + |\sqrt{3} - 5| + |2 - \sqrt{3}|[/tex]
[tex]2\sqrt{3} - 3 = \sqrt{12} - \sqrt{9} > 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]
[tex]b = 2\sqrt{3} - 3 + 5 - \sqrt{3} + 2 - \sqrt{3} = (2\sqrt{3} - 2\sqrt{3}) + (7 - 3) = 4[/tex]
Așadar, a = 6 și b = 4
a > b
