a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne1\\x>0\end{matrix}\right.\)
b)
\(D=\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\dfrac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(1-\sqrt{x}+x-\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)
c)
Giả sử \(D>\dfrac{-2}{\sqrt{x}}\)
\(\Rightarrow\sqrt{x}-1>-\dfrac{2}{\sqrt{x}}\Leftrightarrow\sqrt{x}-1+\dfrac{2}{\sqrt{x}}>0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}+2}{\sqrt{x}}>0\Leftrightarrow x-\sqrt{x}+2>0\Leftrightarrow\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{7}{4}>0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\)(luôn đúng)
