9.152. $\frac{1-\log _{4} x}{1+\log _{2} x} \leq \frac{1}{2}$.
Solution.

Let's switch to base 2.

$$
\begin{aligned}
& \text { We have } \frac{1-\frac{1}{2} \log _{2} x}{1+\log _{2} x}-\frac{1}{2} \leq 0, \frac{2-\log _{2} x-1-\log _{2} x}{2\left(1+\log _{2} x\right)} \leq 0, \\
& \frac{1-2 \log _{2} x}{1+\log _{2} x} \leq 0, \frac{\log _{2} x-\frac{1}{2}}{\log _{2} x+1} \geq 0 .
\end{aligned}
$$

Multiplying both sides of this inequality by the square of the denominator, we get

$$
\left\{\begin{array} { l } 
{ ( \operatorname { l o g } _ { 2 } x - \frac { 1 } { 2 } ) ( \operatorname { l o g } _ { 2 } x + 1 ) \geq 0 , } \\
{ \operatorname { l o g } _ { 2 } x + 1 \neq 0 }
\end{array} \Leftrightarrow \left[\begin{array} { l } 
{ \operatorname { l o g } _ { 2 } x \geq \frac { 1 } { 2 } , } \\
{ \operatorname { l o g } _ { 2 } x < - 1 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x \geq \sqrt{2} \\
0<x<\frac{1}{2}
\end{array}\right.\right.\right.
$$

Answer: $x \in\left(0 ; \frac{1}{2}\right) \cup[\sqrt{2} ; \infty)$.