To find the coefficient of \(x\) in the expansion of \(\left( \frac{2}{x} - 3x \right)^5\), we use the binomial theorem:
\(\left( \frac{2}{x} - 3x \right)^5 = \sum_{k=0}^{5} \binom{5}{k} \left( \frac{2}{x} \right)^{5-k} (-3x)^k\)
We need the term where the power of \(x\) is 1. The general term is:
\(\binom{5}{k} \left( \frac{2}{x} \right)^{5-k} (-3x)^k = \binom{5}{k} \cdot \frac{2^{5-k}}{x^{5-k}} \cdot (-3)^k \cdot x^k\)
The power of \(x\) in this term is \(k - (5-k) = 2k - 5\). We set \(2k - 5 = 1\) to find \(k\):
\(2k - 5 = 1\)
\(2k = 6\)
\(k = 3\)
Substitute \(k = 3\) into the general term:
\(\binom{5}{3} \cdot \left( \frac{2}{x} \right)^{2} \cdot (-3x)^3\)
\(= 10 \cdot \frac{4}{x^2} \cdot (-27x^3)\)
\(= 10 \cdot 4 \cdot (-27) \cdot x\)
\(= -1080x\)
Thus, the coefficient of \(x\) is \(-1080\).
