Question

Cho A = \(\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\) Tìm GTLN của P = \(A-9\sqrt{x}\)

Answer 1

\(A=\left(\frac{1}{x-\sqrt{x}}+\frac{\sqrt{x}}{x-\sqrt{x}}\right)\cdot\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)

\(A=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)

\(A=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\)

\(\Rightarrow P=1-\left(\frac{1}{\sqrt{x}}-9\sqrt{x}\right)\)

\(\le1-2\cdot\sqrt{\frac{1}{\sqrt{x}}\cdot9\sqrt{x}}=1-6=-5\)

Dấu "=" \(\Leftrightarrow\frac{1}{\sqrt{x}}=9\sqrt{x}\Leftrightarrow x=\frac{1}{9}\)

Vậy Max P = -5 <=> x = 1/9