9.202. $x^{\log _{2} x}+16 x^{-\log _{2} x}<17$.
Solution.

Domain of definition: $0<x \neq 1$.

Rewrite the given inequality as $x^{\log _{2} x}+\frac{16}{x^{\log _{2} x}}-17<0 \Leftrightarrow$ $\Leftrightarrow x^{2 \log _{2} x}-17 x^{\log _{2} x}+16<0$. Solving this inequality as a quadratic in terms of $x^{\log _{2} x}$, we get $1<x^{\log _{2} x}<16$. Taking the logarithm base 2, we have the inequality $0<\log _{2}^{2} x<4$, which is equivalent to the system of inequalities $\left\{\begin{array}{l}-2<\log _{2} x<2, \\ \log _{2} x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{1}{4}<x<4, \\ x \neq 1 .\end{array}\right.\right.$ Answer: $x \in\left(\frac{1}{4} ; 1\right) \cup(1 ; 4)$.