a) Ta có: \(Q=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{x+\sqrt{x}+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2\sqrt{x}+2-2+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+2\sqrt{x}}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-1}\)
b) Để \(Q=-\frac{1}{2}\) thì \(\frac{\sqrt{x}}{\sqrt{x}-1}=-\frac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}=-\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow2\sqrt{x}=-\sqrt{x}+1\)
\(\Leftrightarrow3\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{3}\)
hay \(x=\frac{1}{9}\)(nhận)
Vậy: Để \(Q=-\frac{1}{2}\) thì \(x=\frac{1}{9}\)
