The equation in the image appears to be:

$\frac{\frac{x}{x - y} - \frac{x}{x + y}}{\frac{2x^3 + 2xy^2}{xy - y^2}} + \frac{y}{2x^2}$

Let me simplify this step-by-step.

Step 1: Simplify the first part $\frac{x}{x - y} - \frac{x}{x + y}$

This is a difference of two fractions. To combine these fractions, we will find a common denominator, which is $(x - y)(x + y)$. So we get:

$\frac{x}{x - y} - \frac{x}{x + y} = \frac{x(x + y) - x(x - y)}{(x - y)(x + y)}$

Simplifying the numerator:

$x(x + y) - x(x - y) = x^2 + xy - x^2 + xy = 2xy$

So, this part becomes:

$\frac{2xy}{(x - y)(x + y)}$

Step 2: Simplify the denominator $\frac{2x^3 + 2xy^2}{xy - y^2}$

Factor out common terms from the numerator:

$\frac{2(x^3 + xy^2)}{xy - y^2} = \frac{2x(x^2 + y^2)}{y(x - y)}$

Step 3: Combine the two fractions

We now have:

$\frac{\frac{2xy}{(x - y)(x + y)}}{\frac{2x(x^2 + y^2)}{y(x - y)}}$

To simplify, multiply by the reciprocal of the denominator:

$= \frac{2xy}{(x - y)(x + y)} \times \frac{y(x - y)}{2x(x^2 + y^2)}$

Cancel out common terms:

• The $2x$ terms cancel out
• The $x - y$ terms cancel out
• The $y$ in the denominator cancels with one $y$ in the numerator

This leaves:

$\frac{y}{x + y}$

Step 4: Add the second term $\frac{y}{2x^2}$

We now add:

$\frac{y}{x + y} + \frac{y}{2x^2}$

We find the common denominator, which is $(x + y)(2x^2)$, and rewrite each fraction:

$\frac{y}{x + y} = \frac{y \cdot 2x^2}{(x + y)(2x^2)} = \frac{2x^2 y}{(x + y)(2x^2)}$ $\frac{y}{2x^2} = \frac{y \cdot (x + y)}{(x + y)(2x^2)} = \frac{y(x + y)}{(x + y)(2x^2)}$

Now we can combine them:

$\frac{2x^2 y + y(x + y)}{(x + y)(2x^2)}$

Simplifying the numerator:

$2x^2 y + y(x + y) = 2x^2 y + xy + y^2$

So, the final expression is:

$\frac{2x^2 y + xy + y^2}{(x + y)(2x^2)}$

This is the simplified version of the original expression.

Would you like more details on any specific part of the process? Here are 5 related questions:

1. How can we simplify complex fractions more efficiently?
2. What are the common tricks for simplifying algebraic expressions with multiple variables?
3. How does factoring help in simplifying expressions?
4. How would you solve similar expressions with higher-degree polynomials?
5. Can this result be interpreted graphically?

Tip: When simplifying expressions involving fractions, look for opportunities to cancel common terms early to reduce complexity.