$8.14 \frac{\sqrt{x}-3}{x-2}>0$.

Solve the inequality:

$8.14 \frac{\sqrt{x}-3}{x-2}>0$.
8.14 Given that $x$ cannot take negative values, we will divide the points $x_{1}=2$ (root of the equation $x-2=0$) and $x_{2}=9$ (root of the equation $\sqrt{x}-3=0$) into intervals, but not the entire number line, only its part $[0, \infty)$. Using the sign curve (Fig. 8.9), we find the solutions to the given inequality.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-272.jpg?height=127&width=584&top_left_y=1609&top_left_x=306)

Fig. 8.9

Answer: $[0,2) \cup(9, \infty)$.

815 According to hint $5^{0}$, this irrational inequality is equivalent to the system

$$
\left\{\begin{array}{l}
x + 61 \geq 0, \\
x + 5 > 0, \\
x + 61 - 5 \\
x^{2}+9 x-36>0
\end{array}\right.
$$

Solving the quadratic equation $x^{2}+9 x-36=0$, we find $x_{1}=-12, x_{2}=3$. We construct the sign curve (in this case, an arc of a parabola) and mark the interval $x>-5$ with an arrow pointing to the right from the point -5 (Fig. 8.10). The solutions to the first and second inequalities of the system coincide on the interval $(3, \infty)$.

Answer: $(3, \infty)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-273.jpg?height=144&width=584&top_left_y=801&top_left_x=306)

Fig. 8.10