a.\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\sqrt{x}-1\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b.\(Q=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)(do \(x\sqrt{x}-1=\sqrt{x}^3-1=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\))
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
c.Để \(Q=\frac{2}{7}\) thì \(\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{2}{7}\)
\(\Leftrightarrow7\sqrt{x}=2x+2\sqrt{x}+2\)
\(\Leftrightarrow2x-5\sqrt{x}+2=0\)
\(\Leftrightarrow2x-4\sqrt{x}-\sqrt{x}+2=0\)
\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\2\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\)(Thỏa mãn đkxđ)
