Question

Tìm GTLN của \(\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\)

Xin c.ơn trước

Answer 1

\(\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=1+\dfrac{5}{\sqrt{x}+1}\) (\(Đk:x\ge0\))
Ta có: \(\sqrt{x}+1\ge1\Rightarrow\dfrac{5}{\sqrt{x}+1}\le5\)

\(\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\le1+5=6\)

\(Max\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=6\Leftrightarrow x=0\)