To find the coefficient of \(\frac{1}{x}\) in the expansion of \(\left( x - \frac{2}{x} \right)^5\), we use the binomial theorem:

\(\left( x - \frac{2}{x} \right)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} \left( -\frac{2}{x} \right)^k\)

We need the term where the power of \(x\) is \(-1\):

\(x^{5-k} \cdot x^{-k} = x^{5-2k}\)

Set \(5-2k = -1\):

\(5 - 2k = -1\)

\(2k = 6\)

\(k = 3\)

Substitute \(k = 3\) into the binomial term:

\(\binom{5}{3} x^{5-3} \left( -\frac{2}{x} \right)^3\)

\(= \binom{5}{3} x^2 \left( -\frac{8}{x^3} \right)\)

\(= \binom{5}{3} \cdot (-8) \cdot x^{-1}\)

\(= 10 \cdot (-8) \cdot x^{-1}\)

\(= -80 \cdot x^{-1}\)

Thus, the coefficient of \(\frac{1}{x}\) is \(-80\).