Cứu pé với ạ BT 10: Cho biểu thức.,$A=(\frac{x-5}{\sqrt{x-2}-\sqrt3}-\frac3{\sqrt{x+1}-\sqrt{x-2}})(\frac{\sqrt{x+1}+\sqrt3}{\sqrt{3x+3}-x-1}+\frac1{\sqrt{x+1}-\sqrt3})$

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{"userId":5406334,"userName":"Chín Hai"}

Điều kiện xác định của biểu thức là $x \geq 0$, $x eq 4$, $x eq 3$

$A = (\frac{x-5}{\sqrt{x}-2-\sqrt{3}} - \frac{3}{\sqrt{x+1}-\sqrt{x-2}}) (\frac{\sqrt{x+1}+\sqrt{3}}{\sqrt{3x+3}-x-1} + \frac{1}{\sqrt{x+1}-\sqrt{3}})$

$A = (\frac{(x-5)(\sqrt{x}-2+\sqrt{3})}{(\sqrt{x}-2)^2-3} - \frac{3(\sqrt{x+1}+\sqrt{x-2})}{x+1-x+2}) (\frac{\sqrt{x+1}+\sqrt{3}}{\sqrt{3(x+1)}-(x+1)} + \frac{\sqrt{x+1}+\sqrt{3}}{x+1-3})$

$A = (\frac{(x-5)(\sqrt{x}-2+\sqrt{3})}{x-4\sqrt{x}+4-3} - \frac{3(\sqrt{x+1}+\sqrt{x-2})}{3}) (\frac{\sqrt{x+1}+\sqrt{3}}{\sqrt{3(x+1)}-(x+1)} + \frac{\sqrt{x+1}+\sqrt{3}}{x-2})$

$A = (\frac{(x-5)(\sqrt{x}-2+\sqrt{3})}{x-4\sqrt{x}+1} - (\sqrt{x+1}+\sqrt{x-2})) ((\sqrt{x+1}+\sqrt{3}) (\frac{1}{\sqrt{3(x+1)}-(x+1)} + \frac{1}{x-2}))$

$A = (\frac{(x-5)(\sqrt{x}-2+\sqrt{3})}{(\sqrt{x}-2-\sqrt{3})(\sqrt{x}-2+\sqrt{3})} - \sqrt{x+1}-\sqrt{x-2}) ((\sqrt{x+1}+\sqrt{3}) (\frac{x-2+\sqrt{3(x+1)}-x-1}{(\sqrt{3(x+1)}-x-1)(x-2)}))$

$A = (\sqrt{x}-2+\sqrt{3} - \sqrt{x+1}-\sqrt{x-2}) ((\sqrt{x+1}+\sqrt{3}) (\frac{-3+\sqrt{3(x+1)}}{(\sqrt{3(x+1)}-x-1)(x-2)}))$