Question

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

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Answer 1

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

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

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

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