Question

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Answer 1

1: Thay x=9 vào A, ta được:

\(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)

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

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

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

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

3: \(P=A\cdot B=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

\(\sqrt{P}< \dfrac{1}{2}\)

=>\(0< =P< \dfrac{1}{4}\)

=>\(\left\{{}\begin{matrix}\dfrac{\sqrt{x}-2}{\sqrt{x}+1}>=0\\\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< \dfrac{1}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=4\\\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-\dfrac{1}{4}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=4\\\dfrac{4\left(\sqrt{x}-2\right)-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=4\\3\sqrt{x}-9< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=4\\0< =x< 9\end{matrix}\right.\Leftrightarrow4< =x< 9\)