Question

simplify
3 root 2/ root 3+root 6 -4 root 3/root 6+root 2 + root 6 /root 2 +root 3

Answer 1

$$\begin{gathered} \frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}{\displaystyle }{\displaystyle +}{\displaystyle }{\displaystyle \sqrt{2}}}+\frac{\sqrt{6}}{\sqrt{2}{\displaystyle }{\displaystyle +}{\displaystyle }{\displaystyle \sqrt{3}}} \\ =\frac{{\displaystyle 3\sqrt{2}}}{{\displaystyle \sqrt{6}+\sqrt{3}}}\times \frac{\sqrt{6}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \sqrt{3}}}{\sqrt{6}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \sqrt{3}}}-\frac{{\displaystyle 4\sqrt{3}}}{{\displaystyle \sqrt{6}+\sqrt{2}}}\times \frac{\sqrt{6}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \sqrt{2}}}{\sqrt{6}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \sqrt{2}}}+\frac{\sqrt{6}}{\sqrt{3}{\displaystyle }{\displaystyle +}{\displaystyle }{\displaystyle \sqrt{2}}}\times \frac{\sqrt{3}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \sqrt{2}}}{\sqrt{3}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \sqrt{2}}} \\ =\frac{3\left[\sqrt{12}-\sqrt{6}\right]}{{\displaystyle {\left(\sqrt{6}\right)}^2}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle {{\displaystyle \left(\sqrt{3}\right)}}^2}}-\frac{4\left[\sqrt{18}-\sqrt{6}\right]}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{2}\right)}^2}+\frac{\sqrt{18}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \sqrt{12}}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{3\left[\sqrt{12}-\sqrt{6}\right]}{6-3}-\frac{4\left[\sqrt{18}-\sqrt{6}\right]}{6-2}+\frac{{\displaystyle \sqrt{18}-\sqrt{12}}}{{\displaystyle 3}-2} \\ =\frac{{\displaystyle 3\left[\sqrt{12}-\sqrt{6}\right]}}{{\displaystyle 3}}-\frac{{\displaystyle 4\left[\sqrt{18}-\sqrt{6}\right]}}{{\displaystyle 4}}+\frac{{\displaystyle \sqrt{18}-\sqrt{12}}}{{\displaystyle 1}} \\ =\sqrt{12}-\sqrt{6}-\sqrt{18}+\sqrt{6}+\sqrt{18}-\sqrt{12} \\ =0 \end{gathered}$$