a: Sửa đề: \(Q=\left(\dfrac{\sqrt{x}}{1-\sqrt{x}}+\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\dfrac{3-\sqrt{x}}{x-1}\)
\(=\dfrac{x+\sqrt{x}+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để Q=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>x=1(loại)