Question

Tìm giá trị nhỏ nhất

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

Answer 1

\(tacó\sqrt{x}+3>3\)

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

<=>-\(\dfrac{3}{\sqrt{x}+3}< \dfrac{3}{3}\)

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

vậy gtnn là 1

Answer 2

Ta có: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+3\ge3\Rightarrow\dfrac{1}{\sqrt{x}+3}\le1\Rightarrow\dfrac{3}{\sqrt{x}+3}\le3\Rightarrow-\dfrac{3}{\sqrt{x}+3}\ge-3\)Dấu bằng xảy ra \(\sqrt{x}+3=3\)\(\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

Vậy MinA = -3 khi x = 0