Question

Answer 1

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

b) Thay \(x=\dfrac{4}{9}\) vào A, ta được:

\(A=2:\left(\dfrac{2}{3}+1\right)=2:\dfrac{5}{3}=\dfrac{6}{5}\)

c) Để \(A=\dfrac{1}{3}\) thì \(\sqrt{x}+1=6\)

hay x=25

Answer 2

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