https://file.aiquickdraw.com/mathbot/file/1726423618447udlovl3j.jpg

Let's go through each of the exercises from the image step by step.

### 1. Prove that:
\[
\frac{1}{\sqrt{1 + \frac{7}{\sqrt{1 + \sqrt{6400}}}}}} = \frac{3}{4}
\]

#### Solution:
First, simplify the inner part of the expression:
- \(\sqrt{6400} = 80\), so:
  \[
  \sqrt{1 + \sqrt{6400}} = \sqrt{1 + 80} = \sqrt{81} = 9
  \]
- Next, we substitute this back into the equation:
  \[
  \frac{1}{\sqrt{1 + \frac{7}{9}}}}
  \]
- The fraction simplifies to:
  \[
  \frac{7}{9}, \text{ so the equation becomes } \frac{1}{\sqrt{1 + \frac{7}{9}}} = \frac{1}{\sqrt{\frac{16}{9}}} = \frac{1}{\frac{4}{3}} = \frac{3}{4}
  \]
Thus, the equation is verified.

---

### 2. Prove that:
\[
\frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}} = 3 - \sqrt{6}
\]

#### Solution:
Multiply the numerator and denominator by \(\sqrt{3} - \sqrt{2}\) to rationalize the denominator:
\[
\frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} (\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}
\]
The denominator becomes:
\[
(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1
\]
So, the expression simplifies to:
\[
\sqrt{3} (\sqrt{3} - \sqrt{2}) = 3 - \sqrt{6}
\]
Thus, the equation is verified.

---

### 3. Consider:
\(A = 4 - 3\sqrt{5}\) and \(B = -1 + 2\sqrt{5}\)

#### Calculations:
- \(A + B\):
  \[
  A + B = (4 - 3\sqrt{5}) + (-1 + 2\sqrt{5}) = 4 - 1 - 3\sqrt{5} + 2\sqrt{5} = 3 - \sqrt{5}
  \]
- \(A - B\):
  \[
  A - B = (4 - 3\sqrt{5}) - (-1 + 2\sqrt{5}) = 4 + 1 - 3\sqrt{5} - 2\sqrt{5} = 5 - 5\sqrt{5}
  \]
- \(A \cdot B\):
  \[
  A \cdot B = (4 - 3\sqrt{5})(-1 + 2\sqrt{5}) = 4(-1) + 4(2\sqrt{5}) - 3\sqrt{5}(-1) + (-3\sqrt{5})(2\sqrt{5})
  \]
  Simplifying each term:
  \[
  = -4 + 8\sqrt{5} + 3\sqrt{5} - 6(5) = -4 + 11\sqrt{5} - 30 = -34 + 11\sqrt{5}
  \]

---

### 4. Factorize \(9 + 4\sqrt{5}\), then deduce a simplification for \(\sqrt{9 + 4\sqrt{5}}\).

We recognize that \(9 + 4\sqrt{5}\) can be written as a binomial squared:
\[
9 + 4\sqrt{5} = \left( \sqrt{5} + 2 \right)^2
\]
So:
\[
\sqrt{9 + 4\sqrt{5}} = \sqrt{\left( \sqrt{5} + 2 \right)^2} = \sqrt{5} + 2
\]

---

### 5. Prove that:
\[
\frac{2}{\sqrt{3} - 1} + \frac{-2}{\sqrt{3} + 1}
\]
is an integer.

#### Solution:
First, simplify each term by rationalizing the denominators.
For the first term:
\[
\frac{2}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{2(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{2(\sqrt{3} + 1)}{3 - 1} = \frac{2(\sqrt{3} + 1)}{2} = \sqrt{3} + 1
\]
For the second term:
\[
\frac{-2}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{-2(\sqrt{3} - 1)}{(\sqrt{3})^2 - 1^2} = \frac{-2(\sqrt{3} - 1)}{2} = -(\sqrt{3} - 1) = -\sqrt{3} + 1
\]
Now, adding the two expressions:
\[
(\sqrt{3} + 1) + (-\sqrt{3} + 1) = 2
\]
Thus, the result is the integer 2.

---

Would you like more details on any of these solutions?

### Related questions:
1. How can you solve similar rationalization problems involving square roots?
2. Can you generalize the technique used in question 5 for any similar pair of fractions?
3. How can you determine when a binomial is a perfect square, as in question 4?
4. What is the significance of simplifying expressions with irrational numbers?
5. How can you approach problems involving expressions with nested square roots?

### Tip:
When simplifying radical expressions, rationalizing the denominator can help make the operations clearer and easier to handle.

1) Prove that: \( \frac{1}{\sqrt{1 + \frac{7}{\sqrt{1 + \sqrt{6400}}}}} = \frac{3}{4} \)
2) Prove that: \( \frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}} = 3 - \sqrt{6} \)
3) Given \( A = 4 - 3\sqrt{5} \) and \( B = -1 + 2\sqrt{5} \), calculate: \( A + B \), \( A - B \), and \( A \cdot B \)
4) Factorize \( 9 + 4\sqrt{5} \), and simplify \( \sqrt{9 + 4\sqrt{5}} \)
5) Prove that: \( \frac{2}{\sqrt{3} - 1} + \frac{-2}{\sqrt{3} + 1} \) is a natural integer.

Learn how to simplify complex radical expressions and apply rationalization techniques. The problems involve simplifying radicals, rationalizing denominators, and factorizing expressions. Suitable for students in Grades 9-12.

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