To rationalize the denominator of:
$\frac{1}{\sqrt{7} - 2}$ we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{7} + 2$:

$\frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2}$

Step 1: Multiply the denominator

$(\sqrt{7} - 2)(\sqrt{7} + 2) = (\sqrt{7})^2 - (2)^2 = 7 - 4 = 3$

Step 2: Multiply the numerator

$1 \times (\sqrt{7} + 2) = \sqrt{7} + 2$

Step 3: Write the final expression

$\frac{\sqrt{7} + 2}{3}$

Thus, the rationalized form of $\frac{1}{\sqrt{7} - 2}$ is:
$\frac{\sqrt{7} + 2}{3}$

Would you like a step-by-step breakdown or further explanation?

Related Questions:

1. How do you rationalize the denominator for $\frac{1}{\sqrt{a} - b}$?
2. What is the purpose of multiplying by the conjugate in rationalization?
3. Can you rationalize $\frac{3}{\sqrt{5} + 1}$?
4. What happens if the denominator is a sum of two square roots, like $\frac{1}{\sqrt{3} + \sqrt{2}}$?
5. How does rationalization help in simplifying expressions in algebra?

Tip:

Always check if the denominator is a difference of squares after multiplying by the conjugate—it should simplify to a rational number.