\(a.\dfrac{\sqrt{x}}{6\sqrt{x}+1}\)
Vì \(0\le\sqrt{x}< 6\sqrt{x}+1\Rightarrow0\le\dfrac{\sqrt{x}}{6\sqrt{x}+1}< 1\)
Để biểu thức có giá trị nguyên => \(\dfrac{\sqrt{x}}{6\sqrt{x}+1}=0\Rightarrow x=0\)
Vậy x=0
\(b.\dfrac{x-2}{\sqrt{x}-3}=\dfrac{x-9+7}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+7}{\sqrt{x}-3}\)
Để biểu thức có giá trị nguyên \(\Rightarrow7⋮\sqrt{x}-3\Rightarrow\sqrt{x}-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{-4;2;4;10\right\}\Rightarrow x\in\left\{4;16;100\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{-4;2;4;10\right\}\Rightarrow x\in\left\{4;16;100\right\}\)
