$8.16 x-3<\sqrt{x-2}$. 

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$8.16 x-3<\sqrt{x-2}$.
8.16 According to the indication $6^{0}$, this irrational inequality is equivalent to the following combination of two systems of inequalities:

$$
\left\{\begin{array} { l } 
{ x - 2 \geq 0 } \\
{ x - 3 \geq 0 } \\
{ x ^ { 2 } - 6 x + 9 < x - 2 }
\end{array} \quad \left\{\begin{array}{l}
x-2 \geq 0 \\
x-3<0
\end{array}\right.\right.
$$

or

$$
\left\{\begin{array} { l } 
{ x \geq 2 } \\
{ x \geq 3 } \\
{ x ^ { 2 } - 7 x + 1 1 < 0 ; }
\end{array} \quad \left\{\begin{array}{l}
x \geq 2 \\
x<3
\end{array}\right.\right.
$$

The solution to the first system is the interval $3 \leq x<\frac{7+\sqrt{5}}{2}$ (Fig. 8.11,a), and the solution to the second system is the interval $2 \leq x<3$ (Fig. 8.11,b). Combining these solutions, we get the final answer.

Answer: $\left[2, \frac{7+\sqrt{5}}{2}\right]$.

b)

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-274.jpg?height=125&width=421&top_left_y=487&top_left_x=428)

Fig. 8.11