图示为 $y = |a - bx|$ 和 $y = |mx - 6|$ 的图像，其中 $a$、$b$ 和 $m$ 是正常数。$y = |a - bx|$ 的图像与 $x$ 轴交于点 $\left(\frac{9}{4}, 0\right)$。 不等式 $|mx - 6| < |a - bx|$ 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 $\frac{kx^2 + 1}{x^2 + (k+1)x + k} > -1,$ 答案用含 $k$ 的式子表示。

1. Express $a$ in terms of $b$ using the $x$-intercept of $y = |a - bx|$.

The graph of $y = |a - bx|$ meets the $x$-axis at $\left(\frac{9}{4}, 0\right)$. Substituting this point gives:
$a - b\left(\frac{9}{4}\right) = 0 \implies a = \frac{9}{4}b$

2. Use the boundary values of the inequality solution to form equations.

The solution set of $|mx - 6| < |a - bx|$ is $\{x : x \in \mathbb{R}, x < 1 \text{ or } x > 3\}$.
The truth value of the inequality changes at the boundary points $x = 1$ and $x = 3$. At these points, the two graphs intersect, so:
At $x = 1$: $|m - 6| = |a - b|$
At $x = 3$: $|3m - 6| = |a - 3b|$

3. Substitute $a$ in terms of $b$ and solve the absolute value equation for $m$.

Since $b$ is a positive constant:
$|a - b| = \left|\frac{9}{4}b - b\right| = \left|\frac{5}{4}b\right| = \frac{5}{4}b$
$|a - 3b| = \left|\frac{9}{4}b - 3b\right| = \left|-\frac{3}{4}b\right| = \frac{3}{4}b$

Substitute these back into the intersection equations:
$|m - 6| = \frac{5}{4}b$
$|3m - 6| = \frac{3}{4}b$

Divide the first equation by the second to eliminate $b$:
$\frac{|m - 6|}{|3m - 6|} = \frac{5/4}{3/4} = \frac{5}{3}$
$3|m - 6| = 5|3m - 6|$

Considering both cases for the absolute values:
Case 1: $3(m - 6) = 5(3m - 6) \implies 3m - 18 = 15m - 30 \implies 12m = 12 \implies m = 1$
Case 2: $3(m - 6) = -5(3m - 6) \implies 3m - 18 = -15m + 30 \implies 18m = 48 \implies m = \frac{8}{3}$

4. Reject the extraneous value of $m$ and calculate $a$ and $b$.

If $m = \frac{8}{3}$, then $\frac{5}{4}b = \left|\frac{8}{3} - 6\right| = \frac{10}{3} \implies b = \frac{8}{3}$.
Then $a = \frac{9}{4}\left(\frac{8}{3}\right) = 6$.
The original inequality would become $\left|\frac{8}{3}x - 6\right| < \left|6 - \frac{8}{3}x\right|$, which simplifies to $\left|\frac{8}{3}x - 6\right| < \left|\frac{8}{3}x - 6\right|$. This has no solution, contradicting the given solution set. Thus, $m = \frac{8}{3}$ is rejected.
(Alternatively, from the diagram, the vertex of $y = |mx - 6|$ is to the right of $\frac{9}{4}$, so $\frac{6}{m} > \frac{9}{4} \implies m < \frac{8}{3}$.)

If $m = 1$, then $\frac{5}{4}b = |1 - 6| = 5 \implies b = 4$.
Then $a = \frac{9}{4}(4) = 9$.
All constants $a=9$, $b=4$, and $m=1$ are positive, as required.

1. Rearrange the inequality to compare with zero and combine into a single algebraic 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 the numerator and the denominator completely.

Numerator:
$(k+1)(x^2 + x + 1)$
Denominator:
$x^2 + kx + x + k = x(x+k) + 1(x+k) = (x+1)(x+k)$
Substituting these back gives:
$\frac{(k+1)(x^2 + x + 1)}{(x+1)(x+k)} > 0$

3. Analyse the signs of the factors in the numerator to simplify the inequality.

Since $k > 1$, we know $k+1 > 0$.
For the quadratic factor $x^2 + x + 1$, completing the square yields:
$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} > 0 \text{ for all real } x$
(Alternatively, its discriminant is $1^2 - 4(1)(1) = -3 < 0$, so it is always positive.)
Since the entire numerator is strictly positive, the inequality requires the denominator to be strictly positive:
$(x+1)(x+k) > 0$

4. Solve the resulting quadratic inequality, considering the relative positions of the roots.

The critical values for the denominator are $x = -1$ and $x = -k$.
Since it is given that $k > 1$, it follows that $-k < -1$.
The parabola $y = (x+1)(x+k)$ opens upwards, so it takes positive values outside the interval $[-k, -1]$.
$x < -k \quad \text{or} \quad x > -1$