$Đk:x>0;x\ne9;x\ne25$
$P=\bigg(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\bigg):\bigg(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\bigg)\\ =\dfrac{\sqrt{x}(\sqrt{x}-3)-2x}{(\sqrt{x}-3)(\sqrt{x}+3)}.\dfrac{\sqrt{x}(\sqrt{x}-3)}{\sqrt{x}-1-2(\sqrt{x}-3)}\\ =\dfrac{-\sqrt{x}(\sqrt{x}+3).\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)(5-\sqrt{x})}\\ =\dfrac{x}{\sqrt{x}-5}$
$P+4=0\Leftrightarrow P=-4\Leftrightarrow \dfrac{x}{\sqrt{x}-5}=-4\\ \Leftrightarrow x+4\sqrt{x}-20=0\\ \left[\begin{matrix}\sqrt{x}=-2-2\sqrt{6}\ (loại)\\ \sqrt{x}=-2+2\sqrt{6}\ (t/m)\end{matrix}\right.$
$\sqrt{x}=-2+2\sqrt{6}\Rightarrow x=28-8\sqrt{6}\ (t/m)$
Vậy $x=28-8\sqrt{6}$ là giá trị cần tìm.
Xin ctlhn
