Răspuns:
a)
[tex]4 \sqrt{7} - 2(2 \sqrt{7} + \sqrt{3} ) + \sqrt{12} = 4 \sqrt{7} - 4 \sqrt{7} - 2 \sqrt{3} + 2 \sqrt{3} = 0[/tex]
b)
[tex]5 \sqrt{15} - 3( \sqrt{15} + 2) + 4(1 - \sqrt{15} ) = 5 \sqrt{15} - 3 \sqrt{15} - 6 + 4 - \sqrt{15} = - 2 + \sqrt{15} [/tex]
c)
[tex]2(2 \sqrt{3} + \sqrt{2} ) - ( \sqrt{8} + 2 \sqrt{12} ) = 4 \sqrt{3} + 2 \sqrt{2} - (2 \sqrt{2} + 2 \times 2 \sqrt{3} ) = 4 \sqrt{3} + 2 \sqrt{2} - (2 \sqrt{2} + 4 \sqrt{3} ) = 4 \sqrt{3} + 2 \sqrt{2} - 2 \sqrt{2} - 4 \sqrt{3} = 0[/tex]
d)
[tex] \sqrt{ 3} (2 \sqrt{2} - 3 \sqrt{3} ) - \sqrt{2} (2 \sqrt{3} - 3 \sqrt{2} ) = 2 \sqrt{6} - 3 \sqrt{9} - 2 \sqrt{6} - 3 \sqrt{4} = - 3 \times 3 - 3 \times 2 = - 9 + 6 = -3[/tex]
