\(P=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}.\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+1\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=x-1\)
b/ \(2x^2+P\le0\Leftrightarrow2x^2+x-1\le0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)\le0\Leftrightarrow2x-1\le0\Rightarrow x\le\frac{1}{2}\)
Kết hợp ĐKXĐ ta được \(0\le x\le\frac{1}{2}\)