Step 1: Understanding the Concept:
For \( n = 6 \) (even), there is one middle term, which is the \( (6/2 + 1) = 4\text{-th} \) term.
: Key Formula or Approach:
\( T_{r+1} = {}^nC_r a^{n-r} b^r \).
Step 2: Detailed Explanation:
Here \( a = \sqrt{x} \), \( b = -1/\sqrt{x} \), and \( n = 6 \).
For the 4th term, \( r = 3 \):
\[ T_4 = {}^6C_3 (\sqrt{x})^{6-3} \left( -\frac{1}{\sqrt{x}} \right)^3 \]
\[ T_4 = {}^6C_3 (x^{1/2})^3 (-1)^3 (x^{-1/2})^3 \]
\[ T_4 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \cdot x^{3/2} \cdot (-1) \cdot x^{-3/2} \]
The powers of \( x \) cancel out (\( x^{3/2} \cdot x^{-3/2} = x^0 = 1 \)):
\[ T_4 = 20 \cdot (-1) = -20 \].
Step 3: Final Answer:
The middle term is -20.