What is |((x^2-5x+4))/((x^2-4))|<= 1 ?

The solution to |((x^2-5x+4))/((x^2-4))| = 5/2

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|((x^2-5x+4))/((x^2-4))|<= 1

\frac{( x ^{2} - 5 x + 4 )}{( x ^{2} - 4 )} \leq 1

0 \leq x \leq \frac{8}{5} or x \geq \frac{5}{2}

Solution steps

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

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

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

\frac{x ^{2} - 5 x + 4}{x ^{2} - 4} \geq - 1 and \frac{x ^{2} - 5 x + 4}{x ^{2} - 4} \leq 1

(x < - 2 or 0 \leq x < 2 or x \geq \frac{5}{2}) and (- 2 < x \leq \frac{8}{5} or x > 2)

Merge Overlapping Intervals

0 \leq x \leq \frac{8}{5} or x \geq \frac{5}{2}

∣ x - 2 ∣ + ∣ x - 3 ∣ < 6

x^{2} - 1 \leq ∣ 2 x - 1 ∣

\frac{x}{( x ^{2} - x - 1 )} < 2

∣ 4 x - 3 ∣ + 2 < 10

\frac{2}{∣ x - 4 ∣} > 1

What is |((x^2-5x+4))/((x^2-4))|<= 1 ?
The solution to |((x^2-5x+4))/((x^2-4))|<= 1 is 0<= x<= 8/5 \lor x>= 5/2