Question

[4root3 divide by 2minus root2] - [30 divided by 4root3 minu 3root 2] - [3root2 divided by 3 plus 2root 3]

Answer 1

$$\begin{gathered} \left[\frac{4\sqrt{3}}{2-\sqrt{2}}\right]-\left[\frac{30}{4\sqrt{3}-3\sqrt{2}}\right]-\left[\frac{3\sqrt{2}}{3+2\sqrt{3}}\right] \\ =\frac{4\sqrt{3}}{2-\sqrt{2}}\times \frac{2+\sqrt{2}}{2+\sqrt{2}}-\frac{30}{4\sqrt{3}-3\sqrt{2}}\times \frac{4\sqrt{3}+3\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}-\frac{3\sqrt{2}}{3+2\sqrt{3}}\times \frac{3-2\sqrt{3}}{3-2\sqrt{3}} \\ =\frac{8\sqrt{3}+4\sqrt{6}}{{\left(2\right)}^2-{\left(\sqrt{2}\right)}^2}-\frac{\left(120\sqrt{3}+90\sqrt{2}\right)}{{\left(4\sqrt{3}\right)}^2-{\left(3\sqrt{2}\right)}^2}-\frac{\left(9\sqrt{2}-6\sqrt{6}\right)}{{\left(3\right)}^2-{\left(2\sqrt{3}\right)}^2} \\ =\frac{8\sqrt{3}+4\sqrt{6}}{4-2}-\frac{\left(120\sqrt{3}+90\sqrt{2}\right)}{48-18}-\frac{\left(9\sqrt{2}-6\sqrt{6}\right)}{9-12} \\ =\frac{8\sqrt{3}+4\sqrt{6}}{2}-\frac{\left(120\sqrt{3}+90\sqrt{2}\right)}{30}+\frac{\left(9\sqrt{2}-6\sqrt{6}\right)}{3} \\ =4\sqrt{3}+2\sqrt{6}-\left(4\sqrt{3}+3\sqrt{2}\right)+\left(3\sqrt{2}-2\sqrt{6}\right) \\ =4\sqrt{3}+2\sqrt{6}-4\sqrt{3}-3\sqrt{2}+3\sqrt{2}-2\sqrt{6} \\ =4\sqrt{3}-4\sqrt{3}+2\sqrt{6}-2\sqrt{6}+3\sqrt{2}-3\sqrt{2} \\ =0 \end{gathered}$$