2) $B =\frac{x+2\sqrt{x}}{x-4}$
3)
Ta có: $P = \frac{A}{B} = \frac{2+\sqrt{x}}{\sqrt{x}}.\frac{x-4}{x+2\sqrt{x}} = \frac{x-4}{x}$
Theo đề bài:
$Px ≤ \frac{3}{2}(\sqrt{x} - 1)$
$⇔ x - 4 ≤ \frac{3}{2}.\sqrt{x} - \frac{3}{2}$
$⇔ x - \frac{3}{2}.\sqrt{x} - \frac{5}{2} ≤ 0$
$⇔ -1 ≤ \sqrt{x} ≤ \frac{5}{2}$
$⇒ 0 ≤ x ≤ \frac{25}{4}$
Vậy $x ∈ (1;2;3;5;6)$
