The problem asks to express the fraction $\frac{3\sqrt{2}-\sqrt{3}}{2\sqrt{3}-\sqrt{2}}$ in the form $\frac{\sqrt{m}}{\sqrt{n}}$, where $m$ and $n$ are whole numbers.

1. Problem Description

The problem asks to express the fraction $\frac{3\sqrt{2}-\sqrt{3}}{2\sqrt{3}-\sqrt{2}}$ in the form $\frac{\sqrt{m}}{\sqrt{n}}$, where $m$ and $n$ are whole numbers.

2. Solution Steps

First, we rationalize the denominator of the given expression by multiplying both the numerator and denominator by the conjugate of the denominator. The conjugate of $2\sqrt{3}-\sqrt{2}$ is $2\sqrt{3}+\sqrt{2}$.

$\frac{3\sqrt{2}-\sqrt{3}}{2\sqrt{3}-\sqrt{2}} = \frac{3\sqrt{2}-\sqrt{3}}{2\sqrt{3}-\sqrt{2}} \cdot \frac{2\sqrt{3}+\sqrt{2}}{2\sqrt{3}+\sqrt{2}}$

$= \frac{(3\sqrt{2}-\sqrt{3})(2\sqrt{3}+\sqrt{2})}{(2\sqrt{3}-\sqrt{2})(2\sqrt{3}+\sqrt{2})}$

$= \frac{3\sqrt{2}(2\sqrt{3}) + 3\sqrt{2}(\sqrt{2}) - \sqrt{3}(2\sqrt{3}) - \sqrt{3}(\sqrt{2})}{(2\sqrt{3})^2 - (\sqrt{2})^2}$

$= \frac{6\sqrt{6} + 6 - 6 - \sqrt{6}}{4(3) - 2}$

$= \frac{5\sqrt{6}}{12-2}$

$= \frac{5\sqrt{6}}{10}$

$= \frac{\sqrt{6}}{2}$

Now we want to express $\frac{\sqrt{6}}{2}$ in the form $\frac{\sqrt{m}}{\sqrt{n}}$. We can write $2$ as $\sqrt{4}$, so

$\frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{\sqrt{4}}$

Now, we can multiply the numerator and the denominator by $\sqrt{\frac{5}{5}}$.

$\frac{\sqrt{6}}{2}=\frac{\sqrt{6}}{\sqrt{4}}=\frac{\sqrt{6}}{\sqrt{4}} \times \frac{\sqrt{\frac{5}{5}}}{\sqrt{\frac{5}{5}}} = \frac{\sqrt{6 \cdot \frac{5}{5}}}{\sqrt{4 \cdot \frac{5}{5}}}$

That's not working. Instead, we can multiply the top and bottom by $\sqrt{\frac{k}{k}}$.

Alternatively, we want to find an equivalent form of the given options.

(a) $\frac{\sqrt{6}}{\sqrt{10}}$

(b) $\frac{\sqrt{150}}{\sqrt{100}} = \frac{\sqrt{25 \cdot 6}}{\sqrt{100}} = \frac{5\sqrt{6}}{10} = \frac{\sqrt{6}}{2}$

(c) $\frac{2\sqrt{6}}{\sqrt{10}}$

(d) $\frac{5\sqrt{6}}{\sqrt{10}}$

3. Final Answer

(b) $\frac{\sqrt{150}}{\sqrt{100}}$