Để \(\dfrac{6}{\sqrt{x}}\in Z\) <=> \(\sqrt{x}\inƯ\left(6\right)\)
mà \(\sqrt{x}\ge0\)
=> \(\sqrt{x}\in\left\{1;2;3;6\right\}\)
\(\Rightarrow x\in\left\{1;4;9;36\right\}\)
Vậy..............
Để \(\dfrac{6}{\sqrt{x}}\)đạt giá trị nguyên thì \(6⋮\sqrt{x}\)
\(\Rightarrow\sqrt{x}\in\left\{1;2;3;6\right\}\)
\(\Rightarrow x\in\left\{1;4;9;36\right\}\)