[Math] Solve absolute value inequation with fraction

Question

I am new to absolute-value inequalities, how can I solve this kind of inequation?

$$ | \frac{x^2-5x+4}{x^2-4} | \le 1 $$

Answer 1

We need to solve the following system $$\frac{x^2-5x+4}{x^2-4}\leq1$$ and

$$\frac{x^2-5x+4}{x^2-4}\geq-1$$ For the first we need to solve $$\frac{x^2-5x+4}{x^2-4}-1\leq0$$ or $$\frac{8-5x}{(x-2)(x+2)}\leq0,$$ which by the intervals method gives $x>2$ or $-2<x\leq1.6$.

The second inequality gives $$\frac{x^2-5x+4}{x^2-4}+1\geq0$$ or $$\frac{2x^2-5x}{x^2-4}\geq0$$ or $$\frac{x(2x-5)}{(x-2)(x+2)}\geq0,$$ which by the intervals method again gives $x\geq2.5$ or $0\leq x<2$ or $x<-2$.

Thus, after solving of this system we'll get the answer: $$[0,1.6]\cup[2.5,+\infty)$$