Răspuns:
[tex]a=[(\frac{1}{\sqrt{6} })^{-2}+\frac{10}{\sqrt{5} }]:2=[(\sqrt{6})^{2}+\frac{10\sqrt{5} }{(\sqrt{5})^{2} }]:2=[6 +2\sqrt{5}] :2=6:2+ 2\sqrt{5}:2=3+\sqrt{5}.~deci~a= 3+\sqrt{5}.\\b=|4-2\sqrt{5}|+\frac{4}{7+3\sqrt{5} }=-( 4-2\sqrt{5})+\frac{4(7-3\sqrt{5})}{(7+3\sqrt{5})(7-3\sqrt{5})} =-4+2\sqrt{5}+\frac{4(7-3\sqrt{5})}{7^{2}-(3\sqrt{5})^{2} }= -4+2\sqrt{5}+\frac{4(7-3\sqrt{5})}{49-45 }= -4+2\sqrt{5}+7-3\sqrt{5}=3-\sqrt{5},~deci~b=3-\sqrt{5}.\\b)~mg=\sqrt{a*b}=\sqrt{(3+\sqrt{5})(3-\sqrt{5})} =[/tex]
[tex]=\sqrt{3^{2}-(\sqrt{5})^{2} }=\sqrt{9-5} =\sqrt{4}=2.[/tex]
Explicație pas cu pas:
