To find the coefficient of \(x\) in the expansion of \(\left( x + \frac{2}{x^2} \right)^7\), we use the binomial theorem:

\(\left( x + \frac{2}{x^2} \right)^7 = \sum_{k=0}^{7} \binom{7}{k} x^{7-k} \left( \frac{2}{x^2} \right)^k\)

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

\(7-k - 2k = 1\)

Solving for \(k\):

\(7 - 3k = 1\)

\(3k = 6\)

\(k = 2\)

Substitute \(k = 2\) into the binomial term:

\(\binom{7}{2} x^{7-2} \left( \frac{2}{x^2} \right)^2 = \binom{7}{2} x^5 \left( \frac{4}{x^4} \right)\)

\(= \binom{7}{2} \cdot 4 \cdot x = 21 \cdot 4 \cdot x = 84x\)

Thus, the coefficient of \(x\) is 84.