Question

Cho A=\(\left(\dfrac{x+8}{x\sqrt{x}+8}-\dfrac{2}{x-2\sqrt{x}+4}\right)\):\(\dfrac{1}{\sqrt{x}-1}\) với x≥0 ; x≠1

Answer 1

\(A=\left(\dfrac{x+8}{x\sqrt{x}+8}-\dfrac{2}{x-2\sqrt{x}+4}\right):\dfrac{1}{\sqrt{x}-1}\left(ĐKXĐ:x\ge0;x\ne1\right)\)

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

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

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

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

Vậy \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\) , với  \(x\ne1;x\ge0\)

Answer 2

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