https://file.aiquickdraw.com/mathbot/file/1730217426471eamffvm4.jpg

The expression in the image is:

\[
\frac{1}{7 + \sqrt{3}} + \frac{1}{7 - \sqrt{3}}
\]

To simplify this, let's go through the steps:

### Step 1: Find a common denominator
The common denominator for the two fractions is \((7 + \sqrt{3})(7 - \sqrt{3})\).

### Step 2: Expand the denominator
We use the difference of squares formula: \((a + b)(a - b) = a^2 - b^2\).

\[
(7 + \sqrt{3})(7 - \sqrt{3}) = 7^2 - (\sqrt{3})^2 = 49 - 3 = 46
\]

So, the common denominator is \(46\).

### Step 3: Rewrite the expression with the common denominator
Now, rewrite each fraction as follows:

\[
\frac{1}{7 + \sqrt{3}} + \frac{1}{7 - \sqrt{3}} = \frac{7 - \sqrt{3}}{46} + \frac{7 + \sqrt{3}}{46}
\]

### Step 4: Combine the fractions
Combine the numerators over the common denominator:

\[
\frac{7 - \sqrt{3} + 7 + \sqrt{3}}{46} = \frac{14}{46} = \frac{7}{23}
\]

### Final Answer
\[
\frac{1}{7 + \sqrt{3}} + \frac{1}{7 - \sqrt{3}} = \frac{7}{23}
\]

Would you like further details or have any questions?

---

Here are some related questions to deepen your understanding:

1. How would you simplify a similar expression like \(\frac{1}{5 + \sqrt{2}} + \frac{1}{5 - \sqrt{2}}\)?
2. What is the importance of rationalizing the denominator in fractions with square roots?
3. How would you expand and simplify an expression like \((a + b)(a - b)\)?
4. What would happen if the denominator had a complex number instead of a square root?
5. Can you find the result of \(\frac{1}{a + \sqrt{b}} + \frac{1}{a - \sqrt{b}}\) for any values of \(a\) and \(b\)?

**Tip:** When working with expressions with square roots, using the difference of squares formula helps in simplifying by eliminating the root in the denominator.

Simplify the expression \( \frac{1}{7 + \sqrt{3}} + \frac{1}{7 - \sqrt{3}} \).

Learn how to simplify the expression \( \frac{1}{7 + \sqrt{3}} + \frac{1}{7 - \sqrt{3}} \) by rationalizing the denominator. This problem covers key algebraic concepts, such as the difference of squares and common denominators. Suitable for students in Grades 9-10.

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