Question

Simplify:

Answer 1

The rationalising factor for $\frac{\sqrt{5}+2}{\sqrt{5}-2}\mathrm{and}\frac{\sqrt{5}-2}{\sqrt{5}+2}$ are $\sqrt{5}+2\mathrm{and}\sqrt{5}-2$ respectively. So
$$\begin{gathered} \frac{\sqrt{5}+2}{\sqrt{5}-2}-\frac{\sqrt{5}-2}{\sqrt{5}+2}=\frac{\left(\sqrt{5}+2\right)\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)} \\ =\frac{{\left(\sqrt{5}+2\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(2\right)}^2}-\frac{{\left(\sqrt{5}-2\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(2\right)}^2}\left[\because \left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =\frac{5+2+4\sqrt{5}}{5-4}-\frac{5+2-4\sqrt{5}}{5-4} \\ =7+4\sqrt{5}-7+4\sqrt{5}=8\sqrt{5} \end{gathered}$$
Hence, $\frac{\sqrt{5}+2}{\sqrt{5}-2}-\frac{\sqrt{5}-2}{\sqrt{5}+2}=8\sqrt{5}$.