Question

Simplify 1/1+2root2 -root 3

Answer 1

$$\begin{gathered} \frac{1}{1+2\sqrt{2}-\sqrt{3}} \\ =\frac{1}{\left(1+2\sqrt{2}\right){\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \sqrt{3}}} \\ =\frac{{\displaystyle 1}}{{\displaystyle \left(1+2\sqrt{2}\right)-\sqrt{3}}}\times \frac{\left(1+2\sqrt{2}\right)+\sqrt{3}}{\left(1+2\sqrt{2}\right)+\sqrt{3}} \\ =\frac{1+2\sqrt{2}+\sqrt{3}}{\left(1+2\sqrt{2}\right){\displaystyle {}^2}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \left(\sqrt{3}\right)}{\displaystyle {}^2}} \\ =\frac{1+2\sqrt{2}+\sqrt{3}}{1+8+4\sqrt{2}-3} \\ =\frac{2\sqrt{2}+\sqrt{3}+1}{6+4\sqrt{2}} \\ =\frac{{\displaystyle 2\sqrt{2}+\sqrt{3}+1}}{{\displaystyle 6+4\sqrt{2}}}\times \frac{6-4\sqrt{2}}{6-4\sqrt{2}} \\ =\frac{12\sqrt{2}+6\sqrt{3}+6-16-4\sqrt{6}-4\sqrt{2}}{36-32} \\ =\frac{8\sqrt{2}+6\sqrt{3}-4\sqrt{6}-10}{4} \\ =\frac{4\sqrt{2}+3\sqrt{3}-2\sqrt{6}-5}{2} \end{gathered}$$