Question

\(B=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
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Answer 1

Ta có: \(B=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)

\(=\dfrac{x-1}{\sqrt{x}}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1+1-\sqrt{x}}\)

\(=\dfrac{x-1}{x-\sqrt{x}}\left(\sqrt{x}+1\right)=\dfrac{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)