Toán khó

B

braga

$P=\sqrt{\dfrac{x^{6}+6x^{3}+9}{x^{2}}}+\sqrt{x^{2}-4x+4}=\sqrt{\left(\dfrac{x^{3}+3}{x}\right)^{2}}+\sqrt{(x-2)^{2}}$

Vậy để $P$ nguyên thì thì $(x-2)\in \mathbb{Z};(x^{3}+3)\vdots x$
$\implies x\in \mathbb{Z} \ \ \ \ \ \ \ \ \ (1)$
$\begin{cases}(x^{3}+3)\vdots x\\x^{3}\vdots x\end{cases}\implies 3\vdots x \ \ \ \ (2)$
Từ $(1)$ và $(2)$
$\implies x\in \left \{1;-1;3;-3 \right \}$
 
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