Question

Tìm GTNN và giá trị tương ứng của x trong:  \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

Answer 1

\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

Có \(\sqrt{x}+2\ge2\forall x\ge0\)

\(\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)\(\Rightarrow-\dfrac{3}{\sqrt{x}+2}\ge-\dfrac{3}{2}\)\(\Leftrightarrow1-\dfrac{3}{\sqrt{x}+2}\ge1-\dfrac{3}{2}=-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\ge-\dfrac{1}{2}\)

Dấu "=" xảy ra khi x=0

Vậy \(min=-\dfrac{1}{2}\Leftrightarrow x=0\)