Find the specified term of the expression in each of the following binomials:
(v) Middle term of `((x^2)/( 4) - (4)/( x^2) )^(10)`

Answer

To find the middle term of the expression \(\left(\frac{x^2}{4} - \frac{4}{x^2}\right)^{10}\), we will follow these steps:

### Step 1: Identify the values of \(n\), \(x\), and \(a\)
The expression can be rewritten as:
\[ \left(\frac{x^2}{4} + \left(-\frac{4}{x^2}\right)\right)^{10} \]
Here, \(n = 10\), \(x = \frac{x^2}{4}\), and \(a = -\frac{4}{x^2}\).

### Step 2: Determine the middle term
Since \(n = 10\) is an even number, the middle term is given by the formula:
\[ T_{r+1} = T_{\frac{n}{2} + 1} = T_{6} \]
Thus, we need to find the 6th term of the expansion.

### Step 3: Use the binomial theorem formula
The general term \(T_r\) in the binomial expansion is given by:
\[ T_r = \binom{n}{r} x^{n-r} a^r \]
For the 6th term, we have \(r = 5\) (since we start counting from \(T_0\)).

### Step 4: Substitute values into the formula
Substituting \(n = 10\), \(r = 5\), \(x = \frac{x^2}{4}\), and \(a = -\frac{4}{x^2}\):
\[ T_6 = \binom{10}{5} \left(\frac{x^2}{4}\right)^{10-5} \left(-\frac{4}{x^2}\right)^5 \]

### Step 5: Calculate \(\binom{10}{5}\)
Using the combination formula:
\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 252 \]

### Step 6: Calculate the powers
Now calculate the powers:
\[ \left(\frac{x^2}{4}\right)^5 = \frac{x^{10}}{4^5} = \frac{x^{10}}{1024} \]
\[ \left(-\frac{4}{x^2}\right)^5 = -\frac{4^5}{x^{10}} = -\frac{1024}{x^{10}} \]

### Step 7: Combine all parts
Now substitute these back into the term:
\[ T_6 = 252 \cdot \frac{x^{10}}{1024} \cdot \left(-\frac{1024}{x^{10}}\right) \]
\[ = 252 \cdot \frac{x^{10} \cdot (-1024)}{1024 \cdot x^{10}} = 252 \cdot (-1) = -252 \]

### Final Answer
Thus, the middle term of the expression \(\left(\frac{x^2}{4} - \frac{4}{x^2}\right)^{10}\) is:
\[ \boxed{-252} \]