Câu 4:
a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
b: Để A là số nguyên thì \(\sqrt{x}+3\inƯ\left(-3\right)\)
=>căn x+3=3
=>x=0