Câu a :
ĐKXĐ \(\left\{{}\begin{matrix}x\ne1\\x\ge0\end{matrix}\right.\)
Câu b :
\(A=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}-x}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}-x}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\)
\(=\sqrt{x}\left(\sqrt{x}-2\right)\)
Câu c :
\(A=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=4\)
