Let's simplify the given expression step by step:

$\frac{\frac{x^2}{9} - \frac{25}{4}}{\frac{4}{x} - \frac{x}{5}}$

Step 1: Simplify the numerator $\frac{x^2}{9} - \frac{25}{4}$

We need a common denominator to subtract these fractions. The least common denominator between 9 and 4 is 36, so we rewrite each fraction as follows:

$\frac{x^2}{9} = \frac{4x^2}{36}, \quad \frac{25}{4} = \frac{225}{36}$

Now, subtract:

$\frac{4x^2}{36} - \frac{225}{36} = \frac{4x^2 - 225}{36}$

Step 2: Simplify the denominator $\frac{4}{x} - \frac{x}{5}$

The least common denominator between $x$ and 5 is $5x$, so we rewrite each term:

$\frac{4}{x} = \frac{20}{5x}, \quad \frac{x}{5} = \frac{x^2}{5x}$

Now, subtract:

$\frac{20}{5x} - \frac{x^2}{5x} = \frac{20 - x^2}{5x}$

Step 3: Combine the two fractions

Now we have:

$\frac{\frac{4x^2 - 225}{36}}{\frac{20 - x^2}{5x}}$

We simplify this by multiplying by the reciprocal of the denominator:

$\frac{4x^2 - 225}{36} \times \frac{5x}{20 - x^2}$

Now multiply the numerators and the denominators:

$\frac{(4x^2 - 225) \cdot 5x}{36 \cdot (20 - x^2)} = \frac{5x(4x^2 - 225)}{36(20 - x^2)}$

Step 4: Factor the terms

We can factor $4x^2 - 225$ as a difference of squares:

$4x^2 - 225 = (2x - 15)(2x + 15)$

Thus, the expression becomes:

$\frac{5x(2x - 15)(2x + 15)}{36(20 - x^2)}$

The denominator $20 - x^2$ can also be factored as a difference of squares:

$20 - x^2 = (\sqrt{20} - x)(\sqrt{20} + x)$

Finally, the simplified form of the expression is:

$\frac{5x(2x - 15)(2x + 15)}{36(\sqrt{20} - x)(\sqrt{20} + x)}$

Would you like further details on any of the steps, or do you have any questions?

Here are 5 related questions for practice:

1. How do you find the least common denominator for two fractions?
2. What is the difference of squares and how is it used to factor expressions?
3. How do you simplify complex rational expressions?
4. What are the steps to rewrite a fraction with a common denominator?
5. How do you multiply and simplify algebraic fractions?

Tip: Always look for common factors that can simplify your expression early to make calculations easier.