Đáp án: $x\in\{16,\dfrac94,\dfrac49,\dfrac1{16},0\}$
Giải thích các bước giải:
Để $\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\in Z$
$\to \dfrac{\sqrt{x}+1+5}{\sqrt{x}+1}\in Z$
$\to 1+\dfrac{5}{\sqrt{x}+1}\in Z$
$\to \dfrac{5}{\sqrt{x}+1}\in Z$
Mà $\sqrt{x}+1\ge 0+1\ge 1$
$\to 0<\dfrac{5}{\sqrt{x}+1}\le 5$
Do $ \dfrac{5}{\sqrt{x}+1}\in Z$
$\to \dfrac{5}{\sqrt{x}+1}\in\{1,2,3,4,5\}$
$\to \sqrt{x}+1\in\{5,\dfrac52,\dfrac53,\dfrac54,1\}$
$\to \sqrt{x}\in\{4,\dfrac32,\dfrac23,\dfrac14,0\}$
$\to x\in\{16,\dfrac94,\dfrac49,\dfrac1{16},0\}$
