We are given that $x = 1 - \sqrt{2}$. The expression we need to evaluate is:

$\frac{x - 1}{x^2}$

Step 1: Calculate $x - 1$

We know $x = 1 - \sqrt{2}$, so:

$x - 1 = (1 - \sqrt{2}) - 1 = -\sqrt{2}$

Step 2: Calculate $x^2$

Now we square $x = 1 - \sqrt{2}$:

$x^2 = (1 - \sqrt{2})^2 = 1^2 - 2 \times 1 \times \sqrt{2} + (\sqrt{2})^2$

Simplifying the above:

$x^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$

Step 3: Evaluate $\frac{x - 1}{x^2}$

Now, we plug in the values we calculated:

$\frac{x - 1}{x^2} = \frac{-\sqrt{2}}{3 - 2\sqrt{2}}$

To simplify, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of $3 - 2\sqrt{2}$, which is $3 + 2\sqrt{2}$:

$\frac{-\sqrt{2}}{3 - 2\sqrt{2}} \times \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}} = \frac{-\sqrt{2}(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})}$

Step 4: Simplify the denominator

The denominator becomes:

$(3 - 2\sqrt{2})(3 + 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1$

Thus, the expression simplifies to:

$\frac{-\sqrt{2}(3 + 2\sqrt{2})}{1} = -\sqrt{2}(3 + 2\sqrt{2})$

Expanding:

$-\sqrt{2}(3 + 2\sqrt{2}) = -3\sqrt{2} - 2(\sqrt{2})^2 = -3\sqrt{2} - 4$

Thus, the value of the expression is:

$\boxed{-4 - 3\sqrt{2}}$

Would you like further clarification or a breakdown of any step?

---

Here are 5 related questions you might find helpful:

1. How do you rationalize denominators when dealing with square roots?
2. Can you show examples of expanding binomials involving square roots?
3. What is the geometric interpretation of square roots in algebra?
4. How do conjugates work in simplifying expressions with radicals?
5. What are other techniques for simplifying complex fractions with roots?

Tip: Always use conjugates when simplifying expressions with square roots in the denominator. It helps remove the radical from the denominator efficiently.