$A$ có nghĩa $⇔\begin{cases}\sqrt[]x \geq 0\\\sqrt[]x-2 \neq 0\\x.\sqrt[]x-8 \neq 0\\\sqrt[]x+2 \neq 0\end{cases}$
$⇔\begin{cases}x ≥0\\\sqrt[]x \neq 2\\\sqrt[]x-2 \neq 0\end{cases}$
$⇔\begin{cases}x ≥0\\x\neq 4\end{cases}$
Rút gọn:
$A=[\dfrac{\sqrt[]x.(x+2.\sqrt[]x+4)}{(\sqrt[]x-2)(x+2.\sqrt[]x+4)}-\dfrac{x.(\sqrt[]x+2)}{(\sqrt[]x-2)(x+2.\sqrt[]x+4)}].(\dfrac{x+2.\sqrt[]x+4}{\sqrt[]x+2})$
$=[\dfrac{x.\sqrt[]x+2.x+4.\sqrt[]x-x.\sqrt[]x-2x}{(\sqrt[]x-2)(x+2.\sqrt[]x+4)}].(\dfrac{x+2.\sqrt[]x+4}{\sqrt[]x+2})$
$=\dfrac{4.\sqrt[]x}{(\sqrt[]x-2)(\sqrt[]x+2)}$
$=\dfrac{4.\sqrt[]x}{x-4}$
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