Đáp án: `A=\frac{\sqrt{x}+3}{\sqrt{x}-2}`
Giải thích các bước giải:
`A=(\frac{3\sqrt{x}+5}{x-\sqrt{x}-2}+\frac{2\sqrt{x}+4}{4-x})÷\frac{\sqrt{x}}{x+\sqrt{x}}`
`=\frac{(3\sqrt{x}+5)(\sqrt{x}+2)-(2\sqrt{x}+4)(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-2)(\sqrt{x}+2)}÷\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}`
`=\frac{(3x+11\sqrt{x}+10)-(2x+6\sqrt{x}+4)}{(\sqrt{x}+1)(\sqrt{x}-2)(\sqrt{x}+2)}÷\frac{1}{\sqrt{x}+1}`
`=\frac{x+5\sqrt{x}+6}{(\sqrt{x}+1)(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+1)`
`=\frac{(\sqrt{x}+2)(\sqrt{x}+3)(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-2)(\sqrt{x}+2)}`
`=\frac{\sqrt{x}+3}{\sqrt{x}-2}`
