Question

Answer 1

ĐK: \(x\ne\pm4;x\ne0\)

Biểu thức trở thành:

\(\left(\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{8x}{x-4}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\\ =\dfrac{4x-8\sqrt{x}-8x}{x-4}:\dfrac{x-1-2\sqrt{x}+4}{x-2\sqrt{x}}\\ =\dfrac{\left(-4x-8\sqrt{x}\right).\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+3\right)}\\ =\dfrac{-4\sqrt{x}\left(x+2\right).\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+3\right)}\\ =\dfrac{-4x}{x-2\sqrt{x}+3}\)