Đáp án:
$A=\dfrac{4\sqrt{x}}{x+4}$
Giải thích các bước giải:
$A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\\ =\left(\dfrac{(\sqrt{x}+1)^2}{(\sqrt{x}+1)(\sqrt{x}-1)}-\dfrac{(\sqrt{x}-1)^2}{(\sqrt{x}+1)(\sqrt{x}-1)}-\dfrac{8\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}\right):\left(\dfrac{\sqrt{x}-x-3}{(\sqrt{x}+1)(\sqrt{x}-1)}-\dfrac{\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}\right)\\ =\dfrac{(\sqrt{x}+1)^2-(\sqrt{x}-1)^2-8\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}:\dfrac{\sqrt{x}-x-3-\sqrt{x}-1}{(\sqrt{x}+1)(\sqrt{x}-1)}\\ =\dfrac{-4\sqrt{x}}{x-1}:\dfrac{-x-4}{x-1}\\ =\dfrac{4\sqrt{x}}{x+4}$
