Consider the set $X$ = $\{ 1,2 \ldots 10 \}$ . Find two disjoint nonempty sunsets $A$ and $B$ of $X$ such that 

a) $A \cup B = X$;

b) $\prod_{x\in A}x$ is divisible by $\prod_{x\in B}x$, where $\prod_{x\in C}x$ is the product of all numbers in $C$;

c) $\frac{ \prod\limits_{x\in A}x}{ \prod\limits_{x\in B}x}$  is as small as possible.
1. **Understanding the Problem:**
   We need to find two disjoint nonempty subsets \(A\) and \(B\) of \(X = \{1, 2, \ldots, 10\}\) such that:
   - \(A \cup B = X\)
   - \(\prod_{x \in A} x\) is divisible by \(\prod_{x \in B} x\)
   - \(\frac{\prod_{x \in A} x}{\prod_{x \in B} x}\) is as small as possible.

2. **Product of All Elements:**
   Since \(A \cup B = X\) and \(A \cap B = \emptyset\), the product of all elements in \(A\) and \(B\) together is:
   \[
   \prod_{x \in A} x \cdot \prod_{x \in B} x = 10!
   \]
   Therefore,
   \[
   \frac{\prod_{x \in A} x}{\prod_{x \in B} x} = \frac{10!}{\left(\prod_{x \in B} x\right)^2}
   \]

3. **Prime Factorization of \(10!\):**
   \[
   10! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7
   \]

4. **Divisibility Condition:**
   For \(\prod_{x \in A} x\) to be divisible by \(\prod_{x \in B} x\), \(\left(\prod_{x \in B} x\right)^2\) must divide \(10!\). This implies that \(\left(\prod_{x \in B} x\right)^2\) must have even powers of all prime factors.

5. **Maximizing \(\left(\prod_{x \in B} x\right)^2\):**
   To minimize \(\frac{10!}{\left(\prod_{x \in B} x\right)^2}\), we need to maximize \(\left(\prod_{x \in B} x\right)^2\). We need to find a subset \(B\) such that \(\left(\prod_{x \in B} x\right)^2\) is as large as possible while still being a divisor of \(10!\).

6. **Finding Suitable Subsets:**
   Let's consider different possible subsets \(B\) and calculate \(\left(\prod_{x \in B} x\right)^2\):

   - **Example 1:**
     \[
     B = \{4, 2, 6, 3, 5\}
     \]
     \[
     \prod_{x \in B} x = 4 \cdot 2 \cdot 6 \cdot 3 \cdot 5 = 720
     \]
     \[
     \left(\prod_{x \in B} x\right)^2 = 720^2 = 2^8 \cdot 3^4 \cdot 5^2
     \]
     This matches the prime factorization of \(10!\).

   - **Example 2:**
     \[
     B = \{8, 2, 9, 5\}
     \]
     \[
     \prod_{x \in B} x = 8 \cdot 2 \cdot 9 \cdot 5 = 720
     \]
     \[
     \left(\prod_{x \in B} x\right)^2 = 720^2 = 2^8 \cdot 3^4 \cdot 5^2
     \]
     This also matches the prime factorization of \(10!\).

   - **Example 3:**
     \[
     B = \{4, 6, 3, 10\}
     \]
     \[
     \prod_{x \in B} x = 4 \cdot 6 \cdot 3 \cdot 10 = 720
     \]
     \[
     \left(\prod_{x \in B} x\right)^2 = 720^2 = 2^8 \cdot 3^4 \cdot 5^2
     \]
     This also matches the prime factorization of \(10!\).

7. **Conclusion:**
   Any of the above sets \(B\) will satisfy the conditions. For each of these sets \(B\), the corresponding set \(A\) will be \(X \setminus B\).

The final answer is \( \boxed{ B = \{4, 2, 6, 3, 5\} } \) or \(B = \{8, 2, 9, 5\}\) or \(B = \{4, 6, 3, 10\}\)