$Đk:x\ge0\\ A=(\dfrac{3\sqrt{x}}{\sqrt{x}-3}+\dfrac{4\sqrt{x}}{\sqrt{x}+3}+\dfrac{7x-3}{9-x}):(\dfrac{2\sqrt{x}-4}{\sqrt{x}-3}-1)\\ =(\dfrac{3\sqrt{x}(\sqrt{x}+3)+4\sqrt{x}(\sqrt{x}-3)-(7x-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}).(\dfrac{\sqrt{x}-3}{2\sqrt{x}-4-\sqrt{x}+3})\\ =(\dfrac{3-3\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}).(\dfrac{\sqrt{x}-3}{\sqrt{x}-1})\\ =\dfrac{-3}{\sqrt{x}+3}$ $Vì\ \sqrt{x}+3\ge3 \Rightarrow A\ge-1\\ \Rightarrow Min_A=-1 khi \sqrt{x}+3=3\Leftrightarrow x=0\ (t/m)$ Vậy $x=0$ là giá trị cần tìm. Cho mình xin ctlhn với ạ
