Rút gọn biểu thức: (căn bậc hai của x/3 + căn bậc hai của x + 2x/9 - x)

* Hướng dẫn giải

$P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{9-x}\right):\left(\frac{\sqrt{x}-1}{x-3\sqrt{x}}-\frac{2}{\sqrt{x}}\right)$

$\begin{array}{l}=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\frac{3\sqrt{x}-x+2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\frac{3\sqrt{x}+x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{5-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\frac{\sqrt{x}\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot \frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-5}\\ =\frac{x}{\sqrt{x}-5}\end{array}$

Vậy $P=\frac{x}{\sqrt{x}-5}$ với $x>0,x\ne 9,x\ne 25$