Question

Tìm GTLN của A = \(\frac{\sqrt{x}-2}{x\sqrt{x}-8}\)

Answer 1

ĐK:x\(\ge0;x\ne4\)

A=\(\frac{\sqrt{x}-2}{x\sqrt{x}-8}=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}=\frac{1}{x+2\sqrt{x}+4}\)

=\(\frac{1}{\left(\sqrt{x}+1\right)^2+3}\)

vì \(x\ge0;x\ne4\Rightarrow\sqrt{x}+1\ge1\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)^2+3\ge4\)

\(\Leftrightarrow\frac{1}{\left(\sqrt{x}+1\right)^2+3}\le\frac{1}{4}\)

hay A\(\le\frac{1}{4}\)dấu = xảy ra khi : x=0(tm)

vậy max A bằng 1/4 tại x=0