Đáp án:
$A=-\dfrac{\sqrt{x}-2}{\sqrt{x}}$
Giải thích các bước giải:
$A=\bigg(\dfrac{1}{\sqrt{x}}-\dfrac{\sqrt{x}-1}{x+2\sqrt{x}}\bigg):\bigg(\dfrac{1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+1}{x-4}\bigg)$
$ĐKXĐ :x>0 ; x\ne4$
$\Leftrightarrow$
$A=\bigg[\dfrac{\sqrt{x}+2-\sqrt{x}+1}{\sqrt{x}.(\sqrt{x}+2)}\bigg]$$:\bigg[\dfrac{\sqrt{x}-2-\sqrt{x}-1}{(\sqrt{x}+2).(\sqrt{x}-2)}\bigg]$ $=\dfrac{3}{\sqrt{x}(\sqrt{x}+2)}.\dfrac{(\sqrt{x}-2).(\sqrt{x}+2)}{-3}=-\dfrac{\sqrt{x}-2}{\sqrt{x}}$
