Question

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

Answer 1

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

Answer 2

chịu TNiên

Answer 3

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