Cho biểu thức P = (căn bậc hai x/ căn bậc hai x +3 - 2(căn bậc hai x +12/ x-9)). căn bậc hai x +5/ căn bậc hai x -8

a) Ta có: $P=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{2\left(\sqrt{x}+12\right)}{x-9}\right).\frac{\sqrt{x}+5}{\sqrt{x}-8}=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{2\left(\sqrt{x}+12\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+5}{\sqrt{x}-8}$

$$\begin{gathered} =\frac{\sqrt{x}\left(\sqrt{x}-3\right)-2\sqrt{x}-24}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+5}{\sqrt{x}-8}=\frac{x-5\sqrt{x}-24}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+5}{\sqrt{x}-8} \\ =\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-8\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+5}{\sqrt{x}-8}=\frac{\sqrt{x}+5}{\sqrt{x}-3} \end{gathered}$$

Vậy $P=\frac{\sqrt{x}+5}{\sqrt{x}-3}$ với $x\ge 0,x\ne 9,x\ne 64$.