Rút gọn: $[\frac{ √x}{ √x +2} + $ $\frac{2 √x}{ √x -2} - $ $\frac{3x+4}{x-4}] . $$\frac{x √x +8}{2}$ ($x≥0 ; x $ $\neq 4 $ )

2 câu trả lời

`((\sqrt{x})/(\sqrt{x}+2)+(2\sqrt{x})/(\sqrt{x}-2)-(3x+4)/(x-4)).(x\sqrt{x}+8)/2` `(x>=0,xne4)`

`=[(\sqrt{x}(\sqrt{x}-2)+2\sqrt{x}(\sqrt{x}+2)-3x-4)/((\sqrt{x}+2)(\sqrt{x}-2))].((\sqrt{x})^3+2^3)/2`

`=[(x-2\sqrt{x}+2x+4\sqrt{x}-3x-4)/((\sqrt{x}-2)(\sqrt{x}+2))].((\sqrt{x}+2)(x-2\sqrt{x}+4))/2`

`=(2\sqrt{x}-4)/((\sqrt{x}-2)(\sqrt{x}+2)).((\sqrt{x}+2)(x-2\sqrt{x}+4))/2`

`=(2(\sqrt{x}-2))/((\sqrt{x}-2)(\sqrt{x}+2)).((\sqrt{x}+2)(x-2\sqrt{x}+4))/(2)`

`=x-2\sqrt{x}+4`

 

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