Đáp án: $x\in\{4,\dfrac14,0\}$
Giải thích các bước giải:
Ta có:
$A=\dfrac{2\sqrt{x}+5}{\sqrt{x}+1}$
$\to A=\dfrac{2(\sqrt{x}+1)+3}{\sqrt{x}+1}$
$\to A=2+\dfrac{3}{\sqrt{x}+1}$
Mà $\sqrt{x}\ge 0\to \dfrac{3}{\sqrt{x}+1}\le\dfrac{3}{0+1}=3$
Để $A\in Z\to 2+\dfrac{3}{\sqrt{x}+1}\in Z$
$\to \dfrac{3}{\sqrt{x}+1}\in Z$
Mà $0<\dfrac{3}{\sqrt{x}+1}\le 3$
$\to \dfrac{3}{\sqrt{x}+1}\in\{1,2,3\}$
$\to \sqrt{x}+1\in\{3,\dfrac32,1\}$
$\to \sqrt{x}\in\{2,\dfrac12,0\}$
$\to x\in\{4,\dfrac14,0\}$
