tìm x sao cho $\frac{cănx}{x.cănx - 3 cănx +3}$ nhận giá trị nguyên

\(\dfrac{\sqrt x}{x\sqrt x-3\sqrt x+3}\) Đk: \(\left\{ \begin{array}{l} x\ge0 \\ x\sqrt x-3\sqrt x+3\ne 0 \end{array} \right .\) Xét \(x\ne 0\) ta có: \(\dfrac{\sqrt x}{x\sqrt x-3\sqrt x+3}=\dfrac{\sqrt x}{\sqrt x(x-3+\dfrac{3}{\sqrt x})}=\dfrac{1}{x-3+\dfrac{3}{\sqrt x}}\) Để có được số nguyên suy ra \(x-3+\dfrac{3}{\sqrt x}=\pm1\) TH1: \(x-3+\dfrac{3}{\sqrt x}=1\) \(\Rightarrow x+\dfrac{3}{\sqrt x}-4=0\) \(\Rightarrow \dfrac{x\sqrt x+3-4\sqrt x}{\sqrt x}=0\) \(\Rightarrow x\sqrt x+3-4\sqrt x=0\) Đặt \(\sqrt x=t(t>0)\) ta có: \(t^3-4t+3=0\) \(\Rightarrow \left\{ \begin{array}{l} t=1 (tm) \\ t=\dfrac{-1+\sqrt {13}}{2}(tm)\\t=\dfrac{-1-\sqrt{13}}{2}(\text{loại})\end{array} \right .\Rightarrow \left\{ \begin{array}{l} x=1 \\ x=\left({\dfrac{-1+\sqrt {13}}{2}}\right)^2 \end{array} \right .\) TH2: \(x-3+\dfrac{3}{\sqrt x}=-1\) \(\Rightarrow x+\dfrac{3}{\sqrt x}-2=0\) \(\Rightarrow \dfrac{x\sqrt x+3-2\sqrt x}{\sqrt x}=0\) \(\Rightarrow x\sqrt x+3-2\sqrt x=0\) Đặt \(\sqrt x=t(t>0)\) ta có: \(t^3-2t+3=0\) \(\Rightarrow t=-1,89... (\text{loại}) \).