a) \(P=\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)
\(P=\left[\dfrac{2\sqrt{x}}{x\left(\sqrt{x}-1\right)+\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right]:\left(\dfrac{x+1+\sqrt{x}}{x+1}\right)\)
\(P=\left[\dfrac{2\sqrt{x}-\left(x+1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]:\left(\dfrac{x+1+\sqrt{x}}{x+1}\right)\)
\(P=\left[\dfrac{-\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]:\left(\dfrac{x+1+\sqrt{x}}{x+1}\right)\)
\(P=\left[\dfrac{-\left(\sqrt{x}-1\right)}{\left(x+1\right)}\right]:\left(\dfrac{x+1+\sqrt{x}}{x+1}\right)\)
\(P=\left[\dfrac{-\left(\sqrt{x}-1\right)}{x+1+\sqrt{x}}\right]\)
b) điều kiện \(x+\sqrt{x}+1\ne0\)
Giả sử P\(\le0\)
\(\Rightarrow\left[\dfrac{-\left(\sqrt{x}-1\right)}{x+1+\sqrt{x}}\right]\le0\)
\(-(\sqrt{x}-1)\le0\)
\(\Rightarrow\sqrt{x}-1\le0\Leftrightarrow\sqrt{x}\le1\Leftrightarrow x\le1\)
Vậy khi x\(\le1\) thì P \(\le0\)
\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+1+\sqrt{x}}{x+1}=P\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right).\dfrac{x+1}{x+1+\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\dfrac{x+1}{x+1+\sqrt{x}}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}.\dfrac{1}{x+1+\sqrt{x}}\)
\(=\dfrac{1-\sqrt{x}}{x+\sqrt{x}+1}\)
b/DKXD:x≥0,x≠1,x≠-1
\(P=\dfrac{1-\sqrt{x}}{x+\sqrt{x}+1}< 0\)
Vi x+can(x)+1>0 , P<0
\(=>1-\sqrt{x}< 0\)
=>x>1
Vay để P<0 thì x>1
