图中显示了 $y = |a - b x|$ 和 $y = |m x - 6|$ 的图像，其中 $a, b$ 和 $m$ 是正常数。$y = |a - b x|$ 的图像与 x 轴相交于点 $\left(\tfrac{9}{4},0\right)$。不等式 $|m x - 6| < |a - b x|$ 的解集为 $\{x : x \in \mathbb{R},\; x < 1 \text{ or } x > 3\}$。求 $a$, $b$ 和 $m$ 的值。 已知 $k$ 是常数且 $k > 1$，解不等式 $\dfrac{kx^2 + 1}{x^2 + (k+1)x + k} > -1$，用 $k$ 表示你的答案。

1. Find the relationship between $a$ and $b$ using the given x-intercept.

Since the graph of $y = |a - bx|$ meets the x-axis at $\left(\frac{9}{4}, 0\right)$, we have:
$|a - b\left(\frac{9}{4}\right)| = 0$
$a - \frac{9}{4}b = 0 \implies a = \frac{9}{4}b$

2. Establish equations based on the boundary values of the inequality.

The boundary points of the inequality solution, $x = 1$ and $x = 3$, correspond to the intersections of the graphs $y = |mx - 6|$ and $y = |a - bx|$.

At $x = 1$:
$|m(1) - 6| = |a - b(1)| \implies |m - 6| = |a - b|$
At $x = 3$:
$|m(3) - 6| = |a - b(3)| \implies |3m - 6| = |a - 3b|$

Substituting $a = \frac{9}{4}b$ into the right-hand sides:
$|a - b| = \left|\frac{9}{4}b - b\right| = \left|\frac{5}{4}b\right| = \frac{5}{4}b \quad \text{(since } b > 0 \text{)}$
$|a - 3b| = \left|\frac{9}{4}b - 3b\right| = \left|-\frac{3}{4}b\right| = \frac{3}{4}b$

So the equations are:

1. $|m - 6| = \frac{5}{4}b$
2. $|3m - 6| = \frac{3}{4}b$

3. Solve the system of absolute value equations for $m$.

Divide equation (2) by equation (1):
$\frac{|3m - 6|}{|m - 6|} = \frac{ \frac{3}{4}b }{ \frac{5}{4}b } = \frac{3}{5}$
$5|3m - 6| = 3|m - 6|$

Removing the absolute values gives two cases:
$5(3m - 6) = 3(m - 6) \quad \text{or} \quad 5(3m - 6) = -3(m - 6)$

Case 1:
$15m - 30 = 3m - 18$
$12m = 12 \implies m = 1$

Case 2:
$15m - 30 = -3m + 18$
$18m = 48 \implies m = \frac{48}{18} = \frac{8}{3}$

4. Determine the valid value of $m$, then find $a$ and $b$.

If $m = \frac{8}{3}$, then substituting into (1) gives $b = \frac{4}{5}\left|\frac{8}{3} - 6\right| = \frac{8}{3}$.
This gives $a = \frac{9}{4}\left(\frac{8}{3}\right) = 6$.
The two given functions would then be $y = |\frac{8}{3}x - 6|$ and $y = |6 - \frac{8}{3}x|$, which are identical graphs. In this case, the inequality $|mx - 6| < |a - bx|$ has no solution, contradicting the given solution set. Thus, we reject $m = \frac{8}{3}$.

Therefore, $m = 1$.
Substitute $m = 1$ into (1):
$|1 - 6| = \frac{5}{4}b \implies 5 = \frac{5}{4}b \implies b = 4$

Substitute $b = 4$ into $a = \frac{9}{4}b$:
$a = \frac{9}{4}(4) = 9$

1. Rearrange the inequality to compare with zero and express as a single fraction.

$\frac{kx^2 + 1}{x^2 + (k+1)x + k} + 1 > 0$
$\frac{kx^2 + 1 + x^2 + (k+1)x + k}{x^2 + (k+1)x + k} > 0$
$\frac{(k+1)x^2 + (k+1)x + (k+1)}{x^2 + (k+1)x + k} > 0$

2. Factorise both the numerator and the denominator.

$\frac{(k+1)(x^2 + x + 1)}{(x+k)(x+1)} > 0$

3. Identify the signs of the factors in the numerator.

Since $k > 1$, we know $k+1 > 0$.
For the quadratic $x^2 + x + 1$, the discriminant is $\Delta = 1^2 - 4(1)(1) = -3$.
Since $\Delta < 0$ and the coefficient of $x^2$ is positive ($1 > 0$), $x^2 + x + 1 > 0$ for all real $x$.
Therefore, the numerator is always strictly positive.

4. Solve the resulting quadratic inequality for the denominator.

Since the numerator is positive, the fraction is greater than zero if and only if the denominator is also positive:
$(x+k)(x+1) > 0$
The critical values are $x = -k$ and $x = -1$.
Given that $k > 1$, it follows that $-k < -1$.
Thus, the solution is:
$x < -k \text{ or } x > -1$