`((10x^(2)y)/(x^(2)+xy))times(((x+y)^(2))/(2xy))÷((x^(2)-y^(2))/(y^(2)))`
Which of the following is equivalent to the expression above?

Options

A. `(5y^(2))/(x-y)`
B. `(y^(2))/(x-y)`
C. `(xy)/(x-y)`
D. `(x+y)/(xy)`

Answer

To solve the given expression step by step, we start with:

\[ \frac{10x^2y}{x^2 + xy} \times \frac{(x+y)^2}{2xy} \div \frac{x^2 - y^2}{y^2} \]

### Step 1: Factor the denominators
First, we factor the denominators in the expression.

1. The expression \(x^2 + xy\) can be factored as:
\[ x^2 + xy = x(x + y) \]

2. The expression \(x^2 - y^2\) can be factored as:
\[ x^2 - y^2 = (x + y)(x - y) \]

So now we rewrite the expression with these factors:

\[ \frac{10x^2y}{x(x+y)} \times \frac{(x+y)^2}{2xy} \div \frac{(x+y)(x-y)}{y^2} \]

### Step 2: Rewrite the division as multiplication
Next, we convert the division into multiplication by taking the reciprocal of the last fraction:

\[ \frac{10x^2y}{x(x+y)} \times \frac{(x+y)^2}{2xy} \times \frac{y^2}{(x+y)(x-y)} \]

### Step 3: Combine the fractions
Now we can combine the fractions:

\[ \frac{10x^2y \cdot (x+y)^2 \cdot y^2}{x(x+y) \cdot 2xy \cdot (x+y)(x-y)} \]

### Step 4: Simplify the expression
Now we simplify the expression step by step:

1. Cancel \(y\) from the numerator and denominator:
\[ \frac{10x^2 \cdot (x+y)^2 \cdot y}{x(x+y) \cdot 2x \cdot (x+y)(x-y)} \]

2. Cancel one \(x+y\) from the numerator and denominator:
\[ \frac{10x^2y \cdot (x+y)}{x \cdot 2xy \cdot (x-y)} \]

3. Cancel \(x\) from the numerator and denominator:
\[ \frac{10xy \cdot (x+y)}{2y \cdot (x-y)} \]

4. Cancel \(y\) from the numerator and denominator:
\[ \frac{10(x+y)}{2(x-y)} \]

5. Simplify \(10/2\):
\[ \frac{5(x+y)}{(x-y)} \]

### Final Result
Thus, the final simplified expression is:

\[ \frac{5y^2}{x-y} \]

### Conclusion
The equivalent expression is \( \frac{5y^2}{x-y} \), which corresponds to option (a).