Cho biểu thức P = (x/ x căn bậc hai x - 4 căn bậc hai x - 6/ 3 căn bậc hai x -6 + 1/ căn bậc hai x +2)

a) Với x > 0, $x\ne 4$ ta có:

$$\begin{gathered} P=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right) \\ =\left(\frac{x}{\sqrt{x}\left(x-4\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right) \\ =\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+2}{6} \\ =\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{6}=\frac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{6}=\frac{-1}{\sqrt{x}-2} \end{gathered}$$

Vậy $P=\frac{-1}{\sqrt{x}-2}$ với x > 0, $x\ne 4$.