A sequence $ (x_n)$ is given as follows: $ x_0,x_1$ are arbitrary positive real numbers, and $ x_{n\plus{}2}\equal{}\frac{1\plus{}x_{n\plus{}1}}{x_n}$ for $ n \ge 0$. Find $ x_{1998}$.
1. Let \( x_0 = a \) and \( x_1 = b \). We start by using the given recursion formula \( x_{n+2} = \frac{1 + x_{n+1}}{x_n} \).

2. Calculate the first few terms of the sequence:
   \[
   x_2 = \frac{1 + x_1}{x_0} = \frac{1 + b}{a}
   \]
   \[
   x_3 = \frac{1 + x_2}{x_1} = \frac{1 + \frac{1 + b}{a}}{b} = \frac{a + b + 1}{ab}
   \]
   \[
   x_4 = \frac{1 + x_3}{x_2} = \frac{1 + \frac{a + b + 1}{ab}}{\frac{1 + b}{a}} = \frac{\frac{ab + a + b + 1}{ab}}{\frac{1 + b}{a}} = \frac{ab + a + b + 1}{ab} \cdot \frac{a}{1 + b} = \frac{a(1 + b) + a + b + 1}{ab(1 + b)} = \frac{a + 1}{b}
   \]
   \[
   x_5 = \frac{1 + x_4}{x_3} = \frac{1 + \frac{a + 1}{b}}{\frac{a + b + 1}{ab}} = \frac{\frac{b + a + 1}{b}}{\frac{a + b + 1}{ab}} = \frac{b + a + 1}{b} \cdot \frac{ab}{a + b + 1} = a
   \]
   \[
   x_6 = \frac{1 + x_5}{x_4} = \frac{1 + a}{\frac{a + 1}{b}} = \frac{1 + a}{a + 1} \cdot b = b
   \]

3. We observe that \( x_5 = x_0 \) and \( x_6 = x_1 \). This indicates that the sequence is periodic with period 5, i.e., \( x_{n+5} = x_n \) for all \( n \ge 0 \).

4. To find \( x_{1998} \), we need to determine the remainder when 1998 is divided by 5:
   \[
   1998 \div 5 = 399 \text{ remainder } 3
   \]
   Thus, \( 1998 \equiv 3 \pmod{5} \).

5. Therefore, \( x_{1998} = x_3 \).

6. From our earlier calculation, we have:
   \[
   x_3 = \frac{a + b + 1}{ab}
   \]

The final answer is \(\boxed{\frac{a + b + 1}{ab}}\).