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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.

a) Find the value of the expression: 1 / (1 + √2) + 1 / (√2 + √3) + 1 / (√3 + √4). b) If x = √3 + √2, determine whether x + 1/x is rational or irrational.

Solve the expression 1 / (1 + √2) + 1 / (√2 + √3) + 1 / (√3 + √4) by rationalizing the denominators. Also, explore whether x = √3 + √2 plus its reciprocal is rational or irrational. Learn about rationalizing fractions and the properties of irrational numbers, suitable for students in Grades 9-11.

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