函数 $f$ 对 $x \ne 2$ 定义为 $f : x \mapsto \dfrac{kx^{2}-5x+3}{x-2}$， 其中 $k$ 是一个实常数。求 $k$ 的取值范围，使得该函数对所有实数 $x, x \ne 2$ 都是单调递增函数。

1. Find the derivative of the function $f(x)$ using the quotient rule.

$f'(x) = \frac{\text{d}}{\text{d}x} \left( \frac{kx^2 - 5x + 3}{x - 2} \right)$
Using the quotient rule:
$f'(x) = \frac{(2kx - 5)(x - 2) - (kx^2 - 5x + 3)(1)}{(x - 2)^2}$
$f'(x) = \frac{2kx^2 - 4kx - 5x + 10 - kx^2 + 5x - 3}{(x - 2)^2}$
$f'(x) = \frac{kx^2 - 4kx + 7}{(x - 2)^2}$

2. State the condition for the function to be increasing.

For $f(x)$ to be an increasing function for all $x \neq 2$, we require $f'(x) \geq 0$ for all $x \neq 2$.
$\frac{kx^2 - 4kx + 7}{(x - 2)^2} \geq 0$
Since $(x - 2)^2 > 0$ for all $x \neq 2$, this inequality holds if and only if the numerator is non-negative:
$kx^2 - 4kx + 7 \geq 0 \quad \text{for all } x \neq 2.$

3. Formulate conditions for the numerator to be consistently non-negative.

If $k = 0$, the numerator becomes $0x^2 - 4(0)x + 7 = 7 \geq 0$, which is valid for all real $x$.
If $k \neq 0$, the quadratic expression must have a positive leading coefficient ($k > 0$) and a non-positive discriminant ($\Delta \leq 0$) so that it is always non-negative:
$\Delta = (-4k)^2 - 4(k)(7) \leq 0$
$16k^2 - 28k \leq 0$
$4k(4k - 7) \leq 0$

4. Solve the inequality to find the final range of $k$.

Solving $4k(4k - 7) \leq 0$ yields:
$0 \leq k \leq \frac{7}{4}$
Combining this with the valid case $k = 0$, the complete range of values for $k$ is:
$0 \leq k \leq \frac{7}{4}$