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We are given the equation:

\[
\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = a + b\sqrt{3}
\]

We need to find the values of \(a\) and \(b\).

### Solution:

1. **Rationalize the denominator**: We begin by multiplying both the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{3} - 1\). This eliminates the square roots from the denominator.

\[
\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)}
\]

2. **Simplify the denominator**: The denominator is a difference of squares:

\[
(\sqrt{3} + 1)(\sqrt{3} - 1) = (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2
\]

So, the equation becomes:

\[
\frac{(\sqrt{3} - 1)^2}{2}
\]

3. **Expand the numerator**: Now, expand \((\sqrt{3} - 1)^2\):

\[
(\sqrt{3} - 1)^2 = (\sqrt{3})^2 - 2(\sqrt{3})(1) + (1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}
\]

Thus, the expression becomes:

\[
\frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}
\]

4. **Compare with the original form**: We now compare this with the form \(a + b\sqrt{3}\). From the expression \(2 - \sqrt{3}\), we see that:

\[
a = 2 \quad \text{and} \quad b = -1
\]

### Final Answer:
\[
a = 2, \quad b = -1
\]

Would you like more details or have any questions?

Here are some related questions for practice:
1. How do you rationalize a denominator involving square roots?
2. What is the formula for the difference of squares?
3. Can you expand \((\sqrt{2} + 1)^2\)?
4. How do you simplify complex fractions with radicals?
5. What are the steps to express an irrational number in terms of \(a + b\sqrt{c}\)?

**Tip**: When rationalizing the denominator, always use the conjugate to eliminate the square roots.

Find the value of a and b in the equation (√3 - 1) / (√3 + 1) = a + b√3.

Solve the equation (√3 - 1)/(√3 + 1) = a + b√3 by rationalizing the denominator and comparing terms. This step-by-step solution involves key algebraic concepts such as radical expressions, rationalization, and the difference of squares theorem. Suitable for students in Grades 9-12.

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