Question

2 /root5+ root3 + 1 / root3 + root2- 3 / root 5+ root2

Answer 1

$$\begin{gathered} \frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}} \\ =\frac{2}{\sqrt{5}+\sqrt{3}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ =\frac{2\sqrt{5}-2\sqrt{3}}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}+\frac{3\sqrt{2}-3\sqrt{5}}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{2\sqrt{5}-2\sqrt{3}}{5-3}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{3\sqrt{2}-3\sqrt{5}}{5-2} \\ =\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{5} \\ =0 \end{gathered}$$