Đáp án:
a) $A=5$
b) $B=\dfrac{\sqrt x+3}{\sqrt x-2}$
c) $x=256$
Giải thích các bước giải:
ĐKXĐ: $x\ge 0;x\ne 4$
a) Với $x=9$
$A=\dfrac{4\sqrt x-7}{\sqrt x-2}=\dfrac{4.\sqrt 9-7}{\sqrt 9-2}=\dfrac{4.3-7}{3-2}=5$
b) $B=\bigg{(}\dfrac{3\sqrt x+4}{x-4}-\dfrac{1}{\sqrt x+2}\bigg{)}:\dfrac{2}{\sqrt x+2}$
$=\bigg{[}\dfrac{3\sqrt x+4}{(\sqrt x-2)(\sqrt x+2)}-\dfrac{\sqrt x-2}{(\sqrt x-2)(\sqrt x+2)}\bigg{]}:\dfrac{2}{\sqrt x+2}$
$=\dfrac{3\sqrt x+4-\sqrt x+2}{(\sqrt x-2)(\sqrt x+2)}:\dfrac{2}{\sqrt x+2}$
$=\dfrac{2\sqrt x+6}{(\sqrt x-2)(\sqrt x+2)}.\dfrac{\sqrt x+2}{2}$
$=\dfrac{\sqrt x+3}{\sqrt x-2}$
c) $\dfrac{A}{B}=\dfrac{4\sqrt x-7}{\sqrt x-2}:\dfrac{\sqrt x+3}{\sqrt x-2}$
$=\dfrac{4\sqrt x-7}{\sqrt x-2}.\dfrac{\sqrt x-2}{\sqrt x+3}$
$=\dfrac{4\sqrt x-7}{\sqrt x+3}$
$=\dfrac{4(\sqrt x+3)-19}{\sqrt x+3}$
$=4-\dfrac{19}{\sqrt x+3}$
Để $\dfrac{A}{B}\in\mathbb Z$
$⇒4-\dfrac{19}{\sqrt x+3}\in\mathbb Z$
$⇒\dfrac{19}{\sqrt x+3}\in\mathbb Z$
$⇒19\,\vdots \sqrt x+3$
$⇒\sqrt x+3\in Ư_{(19)}$
$⇒\sqrt x+3\in \{±1;±19\}$
Mà $\sqrt x+3>0$
$⇒\sqrt x+3\in\{1;19\}$
TH1: $\sqrt x+3=1$
$⇒\sqrt x=-2\,(L)$
TH2: $\sqrt x+3=19$
$⇒\sqrt x=16$
$⇒x=256$
Vậy $\dfrac{A}{B}\in\mathbb Z$ khi $x=256$.
