Đáp án: `P=\frac{1}{\sqrt{x}-2}(x≥0;x\ne4;x\ne9)`
Giải thích các bước giải:
`P=(\frac{3\sqrt{x}+6}{x-4}+\frac{\sqrt{x}}{\sqrt{x}-2})÷\frac{x-9}{\sqrt{x}-3}`
`=[\frac{3(\sqrt{x}+2)}{x-4}+\frac{\sqrt{x}}{\sqrt{x}-2}]÷\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}-3}`
`=[\frac{3(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{\sqrt{x}}{\sqrt{x}-2}]÷\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}-3}`
`=[\frac{3}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}-2}]÷(\sqrt{x}+3)`
`=\frac{\sqrt{x}+3}{\sqrt{x}-2}.\frac{1}{\sqrt{x}+3}`
`=\frac{1}{\sqrt{x}-2}`
