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We are given the function \( f(x) \):

\[
f(x) = 2 - \frac{(x - 5k)(x - k)}{x^2 - 3kx + 2k^2}, \quad x \geq 0
\]

We are asked to show that:

\[
f(x) = \frac{x + k}{x - 2k}
\]

### Step-by-step approach:

#### Step 1: Simplify the second term

The function \( f(x) \) has the following form:

\[
f(x) = 2 - \frac{(x - 5k)(x - k)}{x^2 - 3kx + 2k^2}
\]

First, simplify the numerator \( (x - 5k)(x - k) \).

\[
(x - 5k)(x - k) = x^2 - kx - 5kx + 5k^2 = x^2 - 6kx + 5k^2
\]

Now, the function becomes:

\[
f(x) = 2 - \frac{x^2 - 6kx + 5k^2}{x^2 - 3kx + 2k^2}
\]

#### Step 2: Combine the two terms

Write the expression as a single fraction:

\[
f(x) = \frac{2(x^2 - 3kx + 2k^2) - (x^2 - 6kx + 5k^2)}{x^2 - 3kx + 2k^2}
\]

Simplify the numerator:

\[
2(x^2 - 3kx + 2k^2) = 2x^2 - 6kx + 4k^2
\]

Now subtract \( (x^2 - 6kx + 5k^2) \):

\[
2x^2 - 6kx + 4k^2 - (x^2 - 6kx + 5k^2) = 2x^2 - 6kx + 4k^2 - x^2 + 6kx - 5k^2
\]

Simplify further:

\[
= x^2 - k^2
\]

Thus, the expression for \( f(x) \) becomes:

\[
f(x) = \frac{x^2 - k^2}{x^2 - 3kx + 2k^2}
\]

#### Step 3: Factor the numerator and denominator

The numerator is a difference of squares:

\[
x^2 - k^2 = (x - k)(x + k)
\]

The denominator can be factored as well:

\[
x^2 - 3kx + 2k^2 = (x - 2k)(x - k)
\]

Thus, the function becomes:

\[
f(x) = \frac{(x - k)(x + k)}{(x - 2k)(x - k)}
\]

#### Step 4: Cancel out the common factor

The term \( (x - k) \) cancels out from the numerator and denominator:

\[
f(x) = \frac{x + k}{x - 2k}
\]

This matches the required result, which completes the proof.

---

Would you like a detailed explanation for any step, or do you have further questions?

Here are 5 questions that might help you explore the problem more:

1. Why is \( k \) given as a negative constant, and how does that impact the domain?
2. What are the restrictions on the function \( f(x) \), particularly at points where the denominator may become zero?
3. Can you explain the difference between simplifying and factoring in algebraic expressions?
4. How would you approach finding the derivative of \( f(x) \) based on the simplified form?
5. What are some other types of functions where a similar method of cancellation occurs?

**Tip**: Always check if expressions can be factored before expanding them—factoring can often simplify the problem significantly.

Given that k is a negative constant and the function f(x) is defined by f(x) = 2 - ((x-5k)(x-k)) / (x^2 - 3kx + 2k^2) for x ≥ 0, show that f(x) = (x+k) / (x-2k).

Learn how to simplify the function f(x) = 2 - (x-5k)(x-k) / (x^2 - 3kx + 2k^2) to show that f(x) = (x+k) / (x-2k). This problem uses key algebraic techniques, such as factoring and simplifying rational expressions. Suitable for students in Grades 10-12.

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