Ta có: $\dfrac{4A}{B}<\dfrac{x}{\sqrt{x}-3}\ (x\geq 0;\ x\neq 1;\ x\neq 9)$$\\$ $\dfrac{4.(\sqrt{x}+2)}{\sqrt{x}-3}:\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\leq\dfrac{x}{\sqrt{x}-3}$ $\dfrac{4.(\sqrt{x}+2)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\leq\dfrac{x}{\sqrt{x}-3}$$\\$ $\dfrac{4(\sqrt{x}-1)}{\sqrt{x}-3}\leq\dfrac{x}{\sqrt{x}-3}$$\\$ $\dfrac{4\sqrt{x}-4}{\sqrt{x}-3}-\dfrac{x}{\sqrt{x}-3}\leq 0$$\\$ $\dfrac{-x+4\sqrt{x}-4}{\sqrt{x}-3}\leq 0$$\\$ $\dfrac{-(x-4\sqrt{x}+4)}{\sqrt{x}-3}\leq 0$$\\$ $\dfrac{-(\sqrt{x}-2)^2}{\sqrt{x}-3}\leq 0$$\\$ Vì $-(\sqrt{x}-2)^2\leq 0\forall x\geq 0$$\\$ Trường hợp $1:$ $-(\sqrt{x}-2)^2=0$$\\$ $\sqrt{x}-2=0$$\\$ $\sqrt{x}=2$$\\$ $x=4$$\\$ Trường hợp $2:$ $\sqrt{x}-3>0$$\\$ $\sqrt{x}>3$$\\$ $x>9$$\\$ Vậy $x=4$ hoặc $x>9$.