求 $k$ 的取值范围，使得 $3x^2 - 5x + k$ is always positive（意为：恒为正数）。 由此，解不等式 $\frac{x^2 - x - 2}{3x^2 - 5x + 4} < 0$。

1. State the condition for the quadratic expression to be always positive

For $3x^2 - 5x + k > 0$ for all real $x$, since the coefficient of $x^2$ is $3 > 0$, the discriminant must be less than zero.
$b^2 - 4ac < 0$

2. Substitute the coefficients into the discriminant and solve for $k$

$(-5)^2 - 4(3)(k) < 0$
$25 - 12k < 0$
$12k > 25$
$k > \frac{25}{12}$

1. Deduce the sign of the denominator using the result from part (a)

From part (a), $3x^2 - 5x + k > 0$ for all real $x$ when $k > \frac{25}{12}$.
Since $4 > \frac{25}{12}$, the denominator $3x^2 - 5x + 4$ is always positive.

2. Simplify the inequality

Since $3x^2 - 5x + 4 > 0$, we can multiply both sides of the inequality by the denominator without changing the inequality sign:
$x^2 - x - 2 < 0$

3. Solve the quadratic inequality

$(x - 2)(x + 1) < 0$
$-1 < x < 2$