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Let's break down the given expression step by step:

The expression is:

\[
\frac{2 \sqrt{48x} - 3 \sqrt{27x}}{\sqrt{243x} - \sqrt{75x}}
\]

### Step 1: Simplify the square roots
We start by simplifying the square roots in both the numerator and the denominator.

- \( \sqrt{48x} = \sqrt{16 \cdot 3 \cdot x} = 4\sqrt{3x} \)
- \( \sqrt{27x} = \sqrt{9 \cdot 3 \cdot x} = 3\sqrt{3x} \)
- \( \sqrt{243x} = \sqrt{81 \cdot 3 \cdot x} = 9\sqrt{3x} \)
- \( \sqrt{75x} = \sqrt{25 \cdot 3 \cdot x} = 5\sqrt{3x} \)

### Step 2: Substitute these simplified square roots back into the expression
The expression now becomes:

\[
\frac{2 \cdot 4 \sqrt{3x} - 3 \cdot 3 \sqrt{3x}}{9 \sqrt{3x} - 5 \sqrt{3x}}
\]

This simplifies to:

\[
\frac{8 \sqrt{3x} - 9 \sqrt{3x}}{9 \sqrt{3x} - 5 \sqrt{3x}}
\]

### Step 3: Simplify the terms
Now simplify both the numerator and the denominator:

- Numerator: \( 8 \sqrt{3x} - 9 \sqrt{3x} = -\sqrt{3x} \)
- Denominator: \( 9 \sqrt{3x} - 5 \sqrt{3x} = 4 \sqrt{3x} \)

So the expression becomes:

\[
\frac{-\sqrt{3x}}{4 \sqrt{3x}}
\]

### Step 4: Cancel out \( \sqrt{3x} \)
Since \( \sqrt{3x} \) appears in both the numerator and the denominator, we can cancel them out:

\[
\frac{-1}{4}
\]

Thus, the expression simplifies to \( -\frac{1}{4} \).

### Final Answer:
The correct answer is \( \boxed{-\frac{1}{4}} \).

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Do you want any further details or have any questions?

Here are some related questions you might find interesting:
1. How do you simplify square roots involving variables?
2. What are the steps to rationalize a denominator?
3. How do you combine like terms in radical expressions?
4. What is the difference between simplifying radicals and rationalizing them?
5. Can you provide an example where the expression involves cube roots instead of square roots?

**Tip:** When simplifying square roots, always look for perfect squares within the radicand to break them down easily.

Simplify the expression: (2√48x - 3√27x) / (√243x - √75x)

Learn how to simplify the expression (2√48x - 3√27x) / (√243x - √75x) with detailed steps, including simplifying square roots and canceling terms. This problem covers key concepts in algebra and radical expressions, suitable for students in Grades 9-11.

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