Đáp án+Giải thích các bước giải:
$B=\dfrac{x}{x-4}-\dfrac{1}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}(x\geq0;x\neq4)$
$=\dfrac{x}{(\sqrt{x}+2)(\sqrt{x}-2)}+\dfrac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}$
$=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{(\sqrt{x}+2)(\sqrt{x}-2)}$
$=\dfrac{x+2\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-2)}$
$=\dfrac{\sqrt{x}}{\sqrt{x}-2}$
Ta có :
$\dfrac{A}{B}=\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{x}=\dfrac{x-4}{x}$
Để $\dfrac{A}{B}.x≤\dfrac{3}{2}(\sqrt{x}-1)$
$⇔\dfrac{x-4}{x}.x≤\dfrac{3}{2}(\sqrt{x}-1)$
$⇔(x-4)≤\dfrac{3}{2}(\sqrt{x}-1)$
$⇔(x-4)-\dfrac{3}{2}(\sqrt{x}-1)≤0$
$⇔\dfrac{2x-8-3\sqrt{x}+3}{2}≤0$
$⇔2x-8-3\sqrt{x}+3≤0$
$⇔2x-3\sqrt{x}-5≤0$
$⇔(\sqrt{x}+1)(2\sqrt{x}-5)≤0$
$⇒2\sqrt{x}-5≤0$(vì$\sqrt{x}+1>0∀x>0$)
$⇔x≤\dfrac{25}{4}$
$⇒0<x≤\dfrac{25}{4}$
$⇒0<x≤6,25$
$⇒x∈{1;2;3;5;6}$
