To find the term independent of x, we use the binomial expansion formula:
\(\left( a + b \right)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).
Here, \(a = 2x\) and \(b = -\frac{1}{4x^2}\), and \(n = 9\).
The general term is \(T_k = \binom{9}{k} (2x)^{9-k} \left(-\frac{1}{4x^2}\right)^k\).
Simplifying, \(T_k = \binom{9}{k} \cdot 2^{9-k} \cdot x^{9-k} \cdot (-1)^k \cdot \frac{1}{4^k} \cdot x^{-2k}\).
\(T_k = \binom{9}{k} \cdot 2^{9-k} \cdot (-1)^k \cdot \frac{1}{4^k} \cdot x^{9-k-2k}\).
\(T_k = \binom{9}{k} \cdot 2^{9-k} \cdot (-1)^k \cdot \frac{1}{4^k} \cdot x^{9-3k}\).
For the term to be independent of x, the power of x must be zero:
\(9 - 3k = 0\).
Solving for \(k\), we get \(k = 3\).
Substitute \(k = 3\) into the expression for \(T_k\):
\(T_3 = \binom{9}{3} \cdot 2^{9-3} \cdot (-1)^3 \cdot \frac{1}{4^3}\).
\(T_3 = \binom{9}{3} \cdot 2^6 \cdot (-1) \cdot \frac{1}{64}\).
\(T_3 = 84 \cdot 64 \cdot (-1) \cdot \frac{1}{64}\).
\(T_3 = -84\).
Thus, the term independent of x is \(-84\).
