\(\dfrac{2x+\sqrt{x}-1}{1-x}-\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1-x\sqrt{x}}\)
\(=\left(2x+\sqrt{x}-1\right)\cdot\left(\dfrac{1}{1-x}-\dfrac{\sqrt{x}}{1-x\sqrt{x}}\right)\)
\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\left(\dfrac{-1}{x-1}+\dfrac{\sqrt{x}}{x\sqrt{x}-1}\right)\)
\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\dfrac{-x-\sqrt{x}-1+x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\left(2\sqrt{x}-1\right)\cdot\dfrac{-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
=>\(P=1+\dfrac{-\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(=1+\dfrac{-\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{x+1}{x+\sqrt{x}+1}\)
b: Để P>2/3 thì P-2/3>0
=>\(\dfrac{x+1}{x+\sqrt{x}+1}-\dfrac{2}{3}>0\)
=>\(3x+3-2\sqrt{x}-2x-2>0\)
=>\(x-2\sqrt{x}+1>0\)(luôn đúng với mọi x>0; x<>1)
