To find the coefficient of \(\frac{1}{x}\) in the expansion, we consider each part separately:

1. For \(\left( kx + \frac{1}{x} \right)^5\), the term that contributes to \(\frac{1}{x}\) is obtained by choosing 4 factors of \(kx\) and 1 factor of \(\frac{1}{x}\). The coefficient is:

\(\binom{5}{1} (kx)^4 \left( \frac{1}{x} \right)^1 = 5 \cdot k^4 \cdot x^3 \cdot \frac{1}{x} = 5k^4 x^2\)

2. For \(\left( 1 - \frac{2}{x} \right)^8\), the term that contributes to \(\frac{1}{x}\) is obtained by choosing 7 factors of 1 and 1 factor of \(-\frac{2}{x}\). The coefficient is:

\(\binom{8}{1} 1^7 \left( -\frac{2}{x} \right)^1 = 8 \cdot (-2) = -16\)

Combining these, the total coefficient of \(\frac{1}{x}\) is:

\(10k^2 - 16\)

We are given that this coefficient is 74:

\(10k^2 - 16 = 74\)

Solving for \(k\):

\(10k^2 = 90\)

\(k^2 = 9\)

\(k = 3\)