Question

Cho A=\(\sqrt{x}-3+\frac{10-x}{\sqrt{x}+3}\) với x≥0
a) Rút gọn A
b) Tìm GTLN của A

Answer 1

\(a.A=\sqrt{x}-3+\frac{10-x}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}+\frac{10-x}{\sqrt{x}+3}=\frac{x-9+10-x}{\sqrt{x}+3}=\frac{1}{\sqrt{x}+3}=\frac{\sqrt{x}-3}{x-9}\)

\(b.\)Ta có: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+3\ge3\forall x\Rightarrow\frac{1}{\sqrt{x}+3}\ge\frac{1}{3}\forall x\)

Vậy \(A_{Min}=\frac{1}{3}\Leftrightarrow x=0\)