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.