Rút gọn bt A=($\frac{3\sqrt{x}+5}{x-\sqrt{x}-2}$+$\frac{2\sqrt{x}+4}{4-x}$):$\frac{\sqrt{x}}{x+\sqrt{x}}$ với x>0;x$\neq$ 4

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1. 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}`

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