Question

\(P=\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\) rút gọn 

Answer 1

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

\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot\dfrac{1+x\sqrt{x}+\sqrt{x}-x\sqrt{x}}{\left(1+x\sqrt{x}\right)\left(1-x\right)}\right]\)

\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\left(2\sqrt{x}-1\right)\cdot\dfrac{1}{\left(x-\sqrt{x}+1\right)\left(1-\sqrt{x}\right)}\right]\)

\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}\cdot\dfrac{\left(x-\sqrt{x}+1\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)