Răspuns:
a) [tex]\frac{6\ }{\sqrt{3} }+2\sqrt{3}-\sqrt{27}=\frac{6\sqrt{3} }{3}+2\sqrt{3}-3\sqrt{3}=2\sqrt{3}+2\sqrt{3}-3\sqrt{3}=4\sqrt{3}=3\sqrt{3}=\sqrt{3}[/tex]
b)[tex]\frac{10}{\sqrt{2} }+\frac{12\sqrt{2} }{\sqrt{6} }-2\sqrt{50} =\frac{10\sqrt{2} }{2} +\frac{12}{\sqrt{3} }-2*5\sqrt{2}=5\sqrt{2} +\frac{12\sqrt{3} }{3}-10\sqrt{2}=\frac{12\sqrt{3} }{3}-5\sqrt{2} =4\sqrt{3}-5\sqrt{2}[/tex]
c)[tex]\frac{5}{\sqrt{2} }+\frac{7}{3\sqrt{2} }-\frac{2\sqrt{3} }{\sqrt{6} } =\frac{5\sqrt{2} }{2}+\frac{7\sqrt{2} }{6} -\frac{2}{\sqrt{2} }=\frac{15\sqrt{2} }{6}+\frac{7\sqrt{2} }{6} -\frac{2\sqrt{2} }{2}=\frac{22\sqrt{2} }{6}-\frac{6\sqrt{2} }{6}=\frac{16\sqrt{2} }{6}=\frac{8\sqrt{2} }{3}[/tex]
