Đáp án: $B=\sqrt{x}$
Giải thích các bước giải:
Ta có:
$B=(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}+2}-\dfrac{8}{x-4}):\dfrac{x+4}{x\sqrt{x}-4\sqrt{x}}$
$\to B=(\dfrac{(\sqrt{x}+2)^2}{(\sqrt{x}-2)(\sqrt{x}+2)}-\dfrac{4(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}-\dfrac{8}{(\sqrt{x}-2)(\sqrt{x}+2)}):\dfrac{x+4}{\sqrt{x}(x-4)}$
$\to B=\dfrac{(\sqrt{x}+2)^2-4(\sqrt{x}-2)-8}{(\sqrt{x}-2)(\sqrt{x}+2)}:\dfrac{x+4}{\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+2)}$
$\to B=\dfrac{(x+4\sqrt{x}+4)-(4\sqrt{x}-8)-8}{(\sqrt{x}-2)(\sqrt{x}+2)}\cdot \dfrac{\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+2)}{x+4}$
$\to B=\dfrac{x+4}{(\sqrt{x}-2)(\sqrt{x}+2)}\cdot \dfrac{\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+2)}{x+4}$
$\to B=\sqrt{x}$
