Question

Fill in the blanks:

$\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}=\mathrm{.............}$

Answer 1

Step 1: Rationalise the denominator of each of the expressions

The rationalising factor of $\sqrt{5}+\sqrt{3},\sqrt{3}+\sqrt{2}$and $\sqrt{5}+\sqrt{2}$ are $\sqrt{5}-\sqrt{3},\sqrt{3}-\sqrt{2}$and $\sqrt{5}-\sqrt{2}$.

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

Step 2: Simplify the above expression.

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

Hence, $\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}=\underset{¯}{0}$