5. In the set of real numbers, solve the inequality:

$$
2|x-2|-\left|\frac{3}{|x-3|}-1\right| \geqslant 1
$$

## Third Year - B Category
5. Above all, we have the condition $x \neq 3$, because the left side is not defined for $x=3$. For $x>3$, the expression $\frac{3}{|x-3|}-1$ reduces to $\frac{3}{x-3}-1$, which is negative for $x-3>3$, i.e., $x>6$, and non-negative for $33$, i.e., $x6$:

The given inequality reduces to $2(x-2)-\left(1-\frac{3}{x-3}\right) \geqslant 1$, i.e., $2 x-6+\frac{3}{x-3} \geqslant 0$, which can also be written as $\frac{2(x-3)^{2}+3}{x-3} \geqslant 0$, and it is clearly always satisfied.

- $3<x \leqslant 6$:

The given inequality reduces to $2(x-2)-\left(\frac{3}{x-3}-1\right) \geqslant 1$, i.e., $2 x-4-\frac{3}{x-3} \geqslant 0$, which can also be written as $\frac{2 x^{2}-10 x+9}{x-3} \geqslant 0$. The function $2 x^{2}-10 x+9$ has zeros at points $x_{1 / 2}=\frac{10 \pm \sqrt{100-72}}{4}=\frac{5 \pm \sqrt{7}}{2}$, so it is non-negative for $x \leqslant \frac{5-\sqrt{7}}{2}$ and $x \geqslant \frac{5+\sqrt{7}}{2}$. Intersecting with the condition $3<x \leqslant 6$, we get the solutions $x \in\left[\frac{5+\sqrt{7}}{2}, 6\right]$.

- $2 \leqslant x<3$:

The given inequality reduces to $2(x-2)-\left(\frac{3}{3-x}-1\right) \geqslant 1$, i.e., $2 x-4-\frac{3}{3-x} \geqslant 0$, which can also be written as $\frac{-2 x^{2}+10 x-15}{3-x} \geqslant 0$. The function $-2 x^{2}+10 x-15$ has a negative discriminant ($100-120=-20<0$), and since the coefficient of the leading term is negative, there are no solutions here.

- $0 \leqslant x<2$:

The given inequality reduces to $2(2-x)-\left(\frac{3}{3-x}-1\right) \geqslant 1$, i.e., $4-2 x-\frac{3}{3-x} \geqslant 0$, which can also be written as $\frac{2 x^{2}-10 x+9}{3-x} \geqslant 0$. The function $2 x^{2}-10 x+9$ is non-negative (as already seen) for $x \leqslant \frac{5-\sqrt{7}}{2}$ and $x \geqslant \frac{5+\sqrt{7}}{2}$. Intersecting with the condition $0 \leqslant x<2$, we get the solutions $x \in\left[0, \frac{5-\sqrt{7}}{2}\right]$.

- $x<0$:

The given inequality reduces to $2(2-x)-\left(1-\frac{3}{3-x}\right) \geqslant 1$, i.e., $2-2 x+\frac{3}{3-x} \geqslant 0$, which can also be written as $\frac{2 x^{2}-8 x+9}{3-x} \geqslant 0$, i.e., $\frac{2(x-2)^{2}+1}{3-x} \geqslant 0$, and it is clearly always satisfied.

Thus, summarizing all, the solution to the inequality is:

$$
x \in\left(-\infty, \frac{5-\sqrt{7}}{2}\right] \cup\left[\frac{5+\sqrt{7}}{2}, \infty\right)
$$

## Third Year - Category B