ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
Để \(A=\frac{1}{3}\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}}=\frac{1}{3}\Leftrightarrow3\sqrt{x}-3=\sqrt{x}\)
\(\Rightarrow2\sqrt{x}=3\Rightarrow\sqrt{x}=\frac{3}{2}\Rightarrow x=\frac{9}{4}\)
c/ \(P=A-9\sqrt{x}=\frac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-\left(\frac{1}{\sqrt{x}}+9\sqrt{x}\right)\)
\(\Rightarrow P\le1-2\sqrt{\frac{9\sqrt{x}}{\sqrt{x}}}=1-6=-5\)
\(\Rightarrow P_{max}=-5\) khi \(\frac{1}{\sqrt{x}}=9\sqrt{x}\Rightarrow x=\frac{1}{9}\)
