函数 $\mathrm{f}$ 和 $\mathrm{g}$ 定义为 $$\begin{aligned} \mathrm{f} : x &\mapsto \frac{x^2 + 2x}{2x^2 - 5x + 2}, \quad x \in \mathbb{R}^+, \ x \ne \frac{1}{2}, 2, \\ \mathrm{g} : x &\mapsto \frac{-3}{2x^2 - 5x + 2}, \quad x \in \mathbb{R}^+, \ x \ne \frac{1}{2}, 2. \end{aligned}$$ 在不使用图形计算器的情况下，求定义域的子集，使得 $\mathrm{f}(x)$ 的值大于或等于 $\mathrm{g}(x)$ 的值。

1. Set up the rational inequality $\mathrm{f}(x) \ge \mathrm{g}(x)$ and combine terms into a single fraction.

$\frac{x^2 + 2x}{2x^2 - 5x + 2} \ge \frac{-3}{2x^2 - 5x + 2}$
$\frac{x^2 + 2x + 3}{2x^2 - 5x + 2} \ge 0$

2. Analyse the numerator to determine its sign.

Completing the square for the numerator:
$x^2 + 2x + 3 = (x+1)^2 + 2$
Since $(x+1)^2 \ge 0$ for all real $x$, the numerator $x^2 + 2x + 3 > 0$ for all real $x$.

3. Formulate and solve the corresponding inequality for the denominator.

Since the numerator is strictly positive, the overall fraction is $\ge 0$ if and only if the denominator is strictly positive (it cannot be zero):
$2x^2 - 5x + 2 > 0$
$(2x - 1)(x - 2) > 0$
Solving this quadratic inequality gives:
$x < \frac{1}{2} \quad \text{or} \quad x > 2$

4. Determine the final subset by intersecting the inequality solution with the given domain.

The function's domain is $x \in \mathbb{R}^+, x \ne \frac{1}{2}, 2$, which means $x > 0$.
Intersecting $x < \frac{1}{2} \text{ or } x > 2$ with $x > 0$ gives:
$0 < x < \frac{1}{2} \quad \text{or} \quad x > 2$
This gives the required subset of the domain.