1: Sửa đề: \(P=\frac{x+2}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
Ta có: \(P=\frac{x+2}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
\(=\frac{x+2-\left(x+\sqrt{x}+1\right)+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2-x-\sqrt{x}-1+x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-2}{x+\sqrt{x}+1}\)
2: \(P<-\frac13\)
=>\(P+\frac13<0\)
=>\(\frac{\sqrt{x}-2}{x+\sqrt{x}+1}+\frac13<0\)
=>\(\frac{3\left(\sqrt{x}-2\right)+x+\sqrt{x}+1}{3\left(x+\sqrt{x}+1\right)}<0\)
=>\(3\sqrt{x}-6+x+\sqrt{x}+1<0\)
=>\(x+4\sqrt{x}-5<0\)
=>\(\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)<0\)
=>\(\sqrt{x}-1<0\)
=>x<1
=>0<=x<1
