Question

Thu gọn giúp với

\(Q=\left(x+\dfrac{1-x\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)^2+\sqrt{x}}\)

Answer 1

\(Q=\left(x-\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}\right)\cdot\dfrac{x+3\sqrt{x}+1}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ Q=\left(x-x-\sqrt{x}-1\right)\cdot\dfrac{x+3\sqrt{x}+1}{\sqrt{x}+1}\\ Q=\dfrac{-\left(\sqrt{x}+1\right)\left(x+3\sqrt{x}+1\right)}{\sqrt{x}+1}=-x-3\sqrt{x}-1\)