The expression to calculate is:

$\left( \frac{\sqrt{3} - 3}{\sqrt{2}} + 4.5\sqrt{15} \right) \cdot \frac{\sqrt{2}}{\sqrt{3}}$

Let’s solve it step by step.

Step 1: Simplify the fraction inside the parentheses

$\frac{\sqrt{3} - 3}{\sqrt{2}}$

This can be rewritten as: $\frac{\sqrt{3}}{\sqrt{2}} - \frac{3}{\sqrt{2}}$

Rationalizing each term (multiply numerator and denominator by $\sqrt{2}$): $\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2}, \quad \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$

So: $\frac{\sqrt{3} - 3}{\sqrt{2}} = \frac{\sqrt{6}}{2} - \frac{3\sqrt{2}}{2}$

Step 2: Add $4.5\sqrt{15}$ to the fraction

The full term inside the parentheses becomes: $\frac{\sqrt{6}}{2} - \frac{3\sqrt{2}}{2} + 4.5\sqrt{15}$

Step 3: Multiply the entire expression by $\frac{\sqrt{2}}{\sqrt{3}}$

The expression becomes: $\left( \frac{\sqrt{6}}{2} - \frac{3\sqrt{2}}{2} + 4.5\sqrt{15} \right) \cdot \frac{\sqrt{2}}{\sqrt{3}}$

Distribute the multiplication: $\frac{\sqrt{6}}{2} \cdot \frac{\sqrt{2}}{\sqrt{3}} - \frac{3\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{\sqrt{3}} + 4.5\sqrt{15} \cdot \frac{\sqrt{2}}{\sqrt{3}}$

Step 4: Simplify each term

1. Simplify the first term: $\frac{\sqrt{6}}{2} \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{12}}{2\sqrt{3}} = \frac{2\sqrt{3}}{2\sqrt{3}} = 1$

2. Simplify the second term: $\frac{3\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{3\sqrt{4}}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$

3. Simplify the third term: $4.5\sqrt{15} \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{4.5\sqrt{30}}{\sqrt{3}} = 4.5\sqrt{10}$

Step 5: Combine everything

The final expression is: $1 - \sqrt{3} + 4.5\sqrt{10}$

This is the simplified result.

Would you like me to expand on any part of this solution or explain any step in detail?

Related Questions:

1. How do you rationalize denominators in fractions involving square roots?
2. What are the steps to simplify terms with square roots in multiplication?
3. Can this problem be solved using a calculator for numerical approximation?
4. What is the importance of simplifying square root expressions in mathematics?
5. How do you handle operations with mixed radicals like $\sqrt{6}$ and $\sqrt{15}$?

Tip:

Always rationalize the denominator to keep expressions in their simplest and most standard form.