To find the coefficient of x in the expansion of \(\left(x^2 - \frac{2}{x}\right)^5\), we use the binomial theorem.
The general term in the expansion is given by:
\(T_k = \binom{5}{k} (x^2)^{5-k} \left(-\frac{2}{x}\right)^k\)
We need the term where the power of x is 1. The power of x in the general term is:
\(2(5-k) - k = 10 - 3k\)
Setting this equal to 1 gives:
\(10 - 3k = 1\)
\(3k = 9\)
\(k = 3\)
Substitute \(k = 3\) into the general term:
\(T_3 = \binom{5}{3} (x^2)^2 \left(-\frac{2}{x}\right)^3\)
\(= 10 \cdot x^4 \cdot \left(-\frac{8}{x^3}\right)\)
\(= 10 \cdot x^4 \cdot \left(-\frac{8}{x^3}\right)\)
\(= 10 \cdot (-8) \cdot x\)
\(= -80x\)
Thus, the coefficient of x is -80.
