Question

Simplify the given surd: $\frac{1}{8}\sqrt{50}+\frac{1}{6}\sqrt{75}-\frac{1}{8}\sqrt{18}-\frac{1}{3}\sqrt{3}$.

Answer 1

The given surds can be simplified as follows:

$\frac{1}{8}\sqrt{50}+\frac{1}{6}\sqrt{75}-\frac{1}{8}\sqrt{18}-\frac{1}{3}\sqrt{3}=\frac{1}{8}\sqrt{25\times 2}+\frac{1}{6}\sqrt{25\times 3}-\frac{1}{8}\sqrt{9\times 2}-\frac{1}{3}\sqrt{3}$

$=(\frac{1}{8}\times 5\sqrt{2})+(\frac{1}{6}\times 5\sqrt{3})-(\frac{1}{8}\times 3\sqrt{2})-\frac{1}{3}\sqrt{3}=\frac{5}{8}\sqrt{2}+\frac{5}{6}\sqrt{3}-\frac{3}{8}\sqrt{2}-\frac{1}{3}\sqrt{3}=\sqrt{2}(\frac{5}{8}-\frac{3}{8})+\sqrt{3}(\frac{5}{6}-\frac{1}{3})$

$=\frac{2}{8}\sqrt{2}+\frac{3}{6}\sqrt{3}=\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{2}$

Hence, $\frac{1}{8}\sqrt{50}+\frac{1}{6}\sqrt{75}-\frac{1}{8}\sqrt{18}-\frac{1}{3}\sqrt{3}=\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{2}$.