\(\dfrac{\sqrt x}{x\sqrt x-3\sqrt x+3}\)
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: \(t^3-2t+3=0\)
\(\Rightarrow t=-1,89... (\text{loại}) \).
