Let's solve the problems shown in the image:

35. (a)

We are asked to find the value of the expression: $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}}$ Each of these terms is a fraction of the form $\frac{1}{a + b}$, which can be simplified by rationalizing the denominator.

First term: $\frac{1}{1 + \sqrt{2}}$

To rationalize this, multiply the numerator and denominator by $1 - \sqrt{2}$: $\frac{1}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{1 - \sqrt{2}}{(1)^2 - (\sqrt{2})^2} = \frac{1 - \sqrt{2}}{1 - 2} = \frac{1 - \sqrt{2}}{-1} = \sqrt{2} - 1$

Second term: $\frac{1}{\sqrt{2} + \sqrt{3}}$

Multiply by $\sqrt{2} - \sqrt{3}$: $\frac{1}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{(\sqrt{2})^2 - (\sqrt{3})^2} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = \frac{\sqrt{2} - \sqrt{3}}{-1} = \sqrt{3} - \sqrt{2}$

Third term: $\frac{1}{\sqrt{3} + \sqrt{4}}$

Multiply by $\sqrt{3} - 2$: $\frac{1}{\sqrt{3} + 2} \times \frac{\sqrt{3} - 2}{\sqrt{3} - 2} = \frac{\sqrt{3} - 2}{(\sqrt{3})^2 - (2)^2} = \frac{\sqrt{3} - 2}{3 - 4} = \frac{\sqrt{3} - 2}{-1} = 2 - \sqrt{3}$

Sum of all terms:

Now add up the results: $(\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3})$ Combine like terms: $\sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} = (-1 + 2) + (\sqrt{2} - \sqrt{2}) + (\sqrt{3} - \sqrt{3}) = 1$

Thus, the value of the expression is: $\boxed{1}$

35. (b)

We are asked to determine whether $x + \frac{1}{x}$ is rational or irrational, where $x = \sqrt{3} + \sqrt{2}$.

Let’s calculate $x + \frac{1}{x}$:

First, find $\frac{1}{x}$: $\frac{1}{x} = \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$

Now, add $x + \frac{1}{x}$: $x + \frac{1}{x} = (\sqrt{3} + \sqrt{2}) + (\sqrt{3} - \sqrt{2}) = 2\sqrt{3}$

Since $\sqrt{3}$ is irrational, multiplying it by 2 gives an irrational number. Therefore, $x + \frac{1}{x}$ is irrational.

Thus, the answer is: $\boxed{\text{Irrational}}$

Let me know if you would like more details or have any further questions!

Here are some related questions to consider:

1. How do you rationalize a denominator when dealing with square roots?
2. Why is $\sqrt{3}$ considered irrational?
3. What are some methods to determine whether a sum of terms is rational or irrational?
4. Can $x + \frac{1}{x}$ ever be rational when $x$ involves square roots?
5. How do operations involving square roots affect the rationality of an expression?

Tip:

When dealing with expressions involving square roots, rationalizing the denominator is often a helpful first step for simplifying complex fractions.