Question

[Nguyễn Giang]

Q = (1 - \(\dfrac{1}{\sqrt{x}}\))\(^2\):(\(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\))
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Answer 1

[Nguyễn Lê Phước Thịnh]

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

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

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

Answer 2

[Nguyễn Giang]

cảm ơn bạn