Question

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\(\dfrac{2\sqrt{x}}{\sqrt{x^3}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\)

Answer 1

Ta có: \(\dfrac{2\sqrt{x}}{\sqrt{x^3}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\)

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

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

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

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

Answer 2

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

Answer 3

Ta có:\(\dfrac{2\sqrt{x}}{\sqrt{x^3}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\)

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

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

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