Sửa đề : \(H=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{\sqrt{x^3}+1}\)
a) Với x > 0, ta có:
\(H=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{\sqrt{x^3}+1}\)
\(=\left[\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}\right)^3+1^3}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\left(\sqrt{x}+1\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Vậy \(H=\dfrac{\sqrt{x}-1}{\sqrt{x}}\).
b) Để \(H=\dfrac{2}{3}\) thì:
\(\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2}{3}\Leftrightarrow3\left(\sqrt{x-1}\right)=2\sqrt{x}\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\) (TMĐK)
Vậy để \(H=\dfrac{2}{3}\) thì x = 9.
c) Để H < 0 thì:
\(\dfrac{\sqrt{x}-1}{\sqrt{x}}< 0\)
vì \(x>0\) nên \(\sqrt{x}>0\).
Do đó để H < 0 thì \(\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
Kết hợp với ĐK: \(0< x< 1\) thì H < 0
Vậy để H < 0 thì 0 < x < 1.