\(\dfrac{\sqrt{x^2-4x}}{3-x}\le2\)
\(\Rightarrow\sqrt{x^2-4x}\le2\left(3-x\right)=6-2x\)
Ta có
\(\left\{{}\begin{matrix}6-2x\ge0\\x^2-4x< \left(6-2x\right)^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\le3\\x^2-4x< 36-24x+4x^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\le3\\3x^2-20x+36>0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\le3\\\forall x\end{matrix}\right.\Rightarrow x\le3\)
Vậy giá trị x thỏa mãn là \(x\le3\)
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