by nathi123 » Tue Jan 29, 2019 7:36 am
A=[tex]\frac{3\sqrt{3}+\sqrt{6}}{\sqrt{6}-2\sqrt{6}+3}=\frac{\sqrt{3}(3+\sqrt{2)}}{3-\sqrt{6}}=\frac{\sqrt{3}(3+\sqrt{2)}}{\sqrt{3}(\sqrt{3}-\sqrt{2})}[/tex]
because [tex]\sqrt{2}*2\sqrt{3}=2\sqrt{2}.\sqrt{3}=2\sqrt{6} ; 3=(\sqrt{3})^{2}[/tex]
[tex]\Rightarrow A=\frac{3+\sqrt{2}}{\sqrt{3}-\sqrt{2}}.\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=3\sqrt{3}+3\sqrt{2}+\sqrt{6}+2[/tex].