Question

Tìm x để Q = \(\dfrac{3\sqrt{x}+15}{\sqrt{x}+3}\)đạt giá trị lớn nhất

Answer 1

ĐKXĐ: \(x\ge0\)

\(Q=\dfrac{3\sqrt{x}+15}{\sqrt{x}+3}=\dfrac{3\left(\sqrt{x}+3\right)+6}{\sqrt{x}+3}=3+\dfrac{6}{\sqrt{x}+3}\)

Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\dfrac{6}{\sqrt{x}+3}\le2\Rightarrow3+\dfrac{6}{\sqrt{x}+3}\le5\)

\(\Rightarrow Q_{max}=5\) khi \(x=0\)

Answer 2

\(Q-5=\dfrac{3\sqrt{x}+15}{\sqrt{x}+3}-5=\dfrac{-2\sqrt{x}}{\sqrt{x}+3}\le0\)

\(\Rightarrow Q\le5\)

\(Q_{maxx}=5\) khi \(\sqrt{x}=0\Leftrightarrow x=0\)