ĐKXĐ: \(x\ge0;x\ne1\)
\(P=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(x=28-6\sqrt{3}=\left(3\sqrt{3}-1\right)^2\Rightarrow\sqrt{x}=3\sqrt{3}-1\)
\(\Rightarrow P=\frac{3\sqrt{3}-1}{28-6\sqrt{3}+3\sqrt{3}-1+1}=\frac{3\sqrt{3}-1}{28-3\sqrt{3}}\)
Ta có: \(P-\frac{1}{3}=\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}=\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\) \(\forall x\ne1\)
\(\Rightarrow P< \frac{1}{3}\)
\(P=\frac{2}{7}\Rightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{2}{7}\Rightarrow7\sqrt{x}=2x+2\sqrt{x}+2\)
\(\Rightarrow2x-5\sqrt{x}+2=0\Rightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=4\end{matrix}\right.\)
Theo chứng minh câu c thì \(P< \frac{1}{3}\) nên P không tồn tại GTLN (dấu = không xảy ra), chỉ tồn tại GTNN \(P=0\) khi \(x=0\)
