Question

Solve the following inequality:
$\left|\frac{x^2-5x+4}{x^2-4}\right|⩽1$

Answer 1

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

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

$\frac{x^2-5x+4}{x^2-4}+1\ge 0$

$\frac{2x^2-5x}{x^2-4}\ge 0$

$\frac{x(2x-5)}{(x-2)(x+2)}\ge 0$

$x\in (-\infty ,-2)\cup (0,2)\cup (\frac{5}{2},\infty )$

And

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

$\frac{x^2-5x+4}{x^2-4}-1\le 0$

$\frac{-5x+8}{x^2-4}\le 0$

$\frac{5x-8}{(x-2)(x+2)}\ge 0$

$x\in (-2,\frac{8}{5})\cup (2,\infty )$

$\therefore x\in (0,\frac{8}{5})\cup (\frac{5}{8},\infty )$