a) ĐK: x > 0; x ≠ 1
\(A=\dfrac{\sqrt{x}+1}{x\sqrt{x}+x\sqrt{x}}:\dfrac{1}{x^2-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}\right)^3}:\dfrac{1}{\sqrt{x}\left[\left(\sqrt{x}\right)^3-1\right]}\)
\(=\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}\right)^3}:\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}\right)^3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{1}\)
\(=\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{2\left(\sqrt{x}\right)^2}\)
\(=\dfrac{\left(x-1\right)\left(x+\sqrt{x}+1\right)}{2x}\)
b) Để A < 0 thì:
\(\dfrac{\left(x-1\right)\left(x+\sqrt{x}+1\right)}{2x}< 0\)
\(\Rightarrow\left(x-1\right)\left(x+\sqrt{x}+1\right)< 0\)
\(\Leftrightarrow x-1< 0\) (vì \(x+\sqrt{x}+1>0\))
\(\Leftrightarrow x< 1\)
Kết hợp điều kiện: \(\left[{}\begin{matrix}x>0\\x< 1\end{matrix}\right.\Rightarrow0< x< 1\)
Vậy để A < 0 thì 0 < x < 1.
