Question

(\(\dfrac{\sqrt{x}-2}{x-1}\) - \(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\) ) x \(\dfrac{x^2-2x+1}{2}\)

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Answer 1

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