Răspuns :
[tex] \sqrt{2} - \sqrt{8} = \sqrt{2} - 2 \times \sqrt{2} = - \sqrt{2} [/tex]
[tex](x - 2)5 + 4 < (3x + 2)3 - 4 \\ 5x - 10 + 4 < 9x + 6 - 4 \\ 5x + 14 < 9x + 2 < = > 14 < 4x + 2 < = > 12 < 4x \\ 3 < x[/tex]
3 <x=>x=4 ,5 ,6........+○○
○○=infinit