Question

Answer 1

Lời giải:

$P=1+\frac{4}{\sqrt{x}+1}$. Để $P$ nguyên thì $\frac{4}{\sqrt{x}+1}$ nguyên.

Ta thấy:

$\sqrt{x}+1>0$ với mọi $x\geq 0$ nên $\frac{4}{\sqrt{x}+1}>0$

Mặt khác:
$\sqrt{x}\geq 0$

$\Rightarrow \sqrt{x}+1\geq 1$

$\Rightarrow \frac{4}{\sqrt{x}+1}\leq 4$

Vậy $0< \frac{4}{\sqrt{x}+1}\leq 4$
$\frac{4}{\sqrt{x}+1}$ nguyên thì $\frac{4}{\sqrt{x}+1}\in \left\{1;2;3;4\right\}$

$\Rightarrow x\in\left\{0; 1; \frac{1}{9}; 9\right\}$