Evaluate $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$, it being given that $\sqrt{5}=2.236$ and $\sqrt{10}=3.162$.
Hint
$\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}=\frac{15}{\sqrt{10}+2 \sqrt{5}+2 \sqrt{10}-\sqrt{5}-4 \sqrt{5}}$
$=\frac{15}{3 \sqrt{10}-3 \sqrt{5}}=\frac{5}{\sqrt{10}-\sqrt{5}}$
$\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}=\frac{15}{\sqrt{10}+2 \sqrt{5}+2 \sqrt{10}-\sqrt{5}-4 \sqrt{5}}$
$=\frac{15}{3 \sqrt{10}-3 \sqrt{5}}$
$=\frac{5}{\sqrt{10}-\sqrt{5}}$
$=\frac{5}{\sqrt{10}-\sqrt{5}} \times \frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}+\sqrt{5}}$
$=\frac{5(\sqrt{10}+\sqrt{5})}{(\sqrt{10})^{2}-(\sqrt{5})^{2}}$
$=\frac{5(\sqrt{10}+\sqrt{5})}{10-5}$
$=\frac{5(\sqrt{10}+\sqrt{5})}{5}$
$=\sqrt{10}+\sqrt{5}$
$=3.162+2.236 \quad$ (given)
$=5.398$
Hence, $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{\overline{40}}-\sqrt{5}-\sqrt{80}}=5.398$.