Find the 7th term in the expansion of `((4x)/5-5/(2x))^9`.

Explanation

To find the 7th term in the expansion of \(\left(\frac{4x}{5} - \frac{5}{2x}\right)^9\), we will use the Binomial Theorem. According to the theorem, the \(r\)th term in the expansion of \((p + q)^n\) is given by:

\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]

### Step-by-Step Solution:

1. **Identify \(p\), \(q\), and \(n\)**:
- Here, \(p = \frac{4x}{5}\), \(q = -\frac{5}{2x}\), and \(n = 9\).

2. **Determine the term we want**:
- We want the 7th term, which corresponds to \(r = 6\) (since \(T_{r+1} = T_7\)).

3. **Write the general term**:
- The general term \(T_{r+1}\) is:
\[ T_{r+1} = \binom{9}{r} \left(\frac{4x}{5}\right)^{9-r} \left(-\frac{5}{2x}\right)^r \]

4. **Substituting \(r = 6\)**:
- Substitute \(r = 6\) into the general term:
\[ T_7 = \binom{9}{6} \left(\frac{4x}{5}\right)^{9-6} \left(-\frac{5}{2x}\right)^6 \]

5. **Calculate \(\binom{9}{6}\)**:
- \(\binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\).

6. **Calculate the powers**:
- \(\left(\frac{4x}{5}\right)^{3} = \frac{(4^3)(x^3)}{5^3} = \frac{64x^3}{125}\)
- \(\left(-\frac{5}{2x}\right)^{6} = \frac{(-5)^6}{(2x)^6} = \frac{15625}{64x^6}\) (since \((-5)^6 = 15625\))

7. **Combine the terms**:
- Now substitute back into the term:
\[ T_7 = 84 \cdot \frac{64x^3}{125} \cdot \frac{15625}{64x^6} \]

8. **Simplify**:
- The \(64\) cancels out:
\[ T_7 = 84 \cdot \frac{15625}{125} \cdot \frac{x^3}{x^6} = 84 \cdot 125 \cdot \frac{1}{x^3} \]
- Calculate \(84 \cdot 125\):
\[ 84 \cdot 125 = 10500 \]

9. **Final Result**:
- Therefore, the 7th term in the expansion is:
\[ T_7 = \frac{10500}{x^3} \]