Solve the inequality: (x2-12)/(x) > 1

The quadratic expression $x^2-x-12$ can be factored as $(x-4)(x+3)$.
So the inequality becomes:

The critical points are the values of $x$ that make the numerator or the denominator equal to zero.
From the numerator: $x-4=0 \implies x=4$ and $x+3=0 \implies x=-3$.
From the denominator: $x=0$.
The critical points are $x=-3, 0, 4$.

The critical points divide the number line into four intervals: $(-\infty, -3)$, $(-3, 0)$, $(0, 4)$, and $(4, \infty)$.

• For $x < -3$ (e.g., $x=-4$): $(x-4)$ is negative. $(x+3)$ is negative. $x$ is negative. The expression is $\frac{(-)(-)}{(-)} = \frac{(+)}{(-)} = (-)$. So, $\frac{(x-4)(x+3)}{x} < 0$.

• For $-3 < x < 0$ (e.g., $x=-1$): $(x-4)$ is negative. $(x+3)$ is positive. $x$ is negative. The expression is $\frac{(-)(+)}{(-)} = \frac{(-)}{(-)} = (+)$. So, $\frac{(x-4)(x+3)}{x} > 0$. This interval is part of the solution.

• For $0 < x < 4$ (e.g., $x=1$): $(x-4)$ is negative. $(x+3)$ is positive. $x$ is positive. The expression is $\frac{(-)(+)}{(+)} = \frac{(-)}{(+)} = (-)$. So, $\frac{(x-4)(x+3)}{x} < 0$.

• For $x > 4$ (e.g., $x=5$): $(x-4)$ is positive. $(x+3)$ is positive. $x$ is positive. The expression is $\frac{(+)(+)}{(+)} = \frac{(+)}{(+)} = (+)$. So, $\frac{(x-4)(x+3)}{x} > 0$. This interval is part of the solution.

The inequality $\frac{(x-4)(x+3)}{x} > 0$ holds for $-3 < x < 0$ or $x > 4$.

The range of values for $x$ is $\boxed{(-3, 0) \cup (4, \infty)}$.