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RÚT GỌN BIỂU THỨC:

9) \(A = \left(\dfrac{1}{1 - \sqrt{x}}+\dfrac{1}{1 + \sqrt{x}}\right) : \left(\dfrac{1}{1 - \sqrt{x}}-\dfrac{1}{1 + \sqrt{x}}\right) + \dfrac{1}{1 - \sqrt{x}} \)

H9
4 tháng 9 2023 lúc 10:38

\(A=\left(\dfrac{1}{1-\sqrt{x}}+\dfrac{1}{1+\sqrt{x}}\right):\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{1}{1+\sqrt{x}}\right)+\dfrac{1}{1-\sqrt{x}}\) (ĐK: \(x>0;x\ne1\))

\(A=\left[\dfrac{1+\sqrt{x}}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}+\dfrac{1-\sqrt{x}}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}\right]:\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{1}{1+\sqrt{x}}\right)+\dfrac{1}{1-\sqrt{x}}\)

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

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

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

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

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

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

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

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

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

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JW
4 tháng 9 2023 lúc 10:42

A = \(\dfrac{1}{\sqrt{x}-x}\) 

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