Question

Tim GTNN cua bieu thuc sau:

Q= \(\frac{-2\sqrt{3x}}{3+x}\) ( voi x≥0, x≠-3)

Answer 1

Đặt \(\sqrt{3x}=t\ge0\Rightarrow x=\frac{t^2}{3}\)

\(Q\left(t\right)=\frac{-2t}{3+\frac{t^2}{3}}=\frac{-6t}{t^2+9}\)

\(\Rightarrow Q'\left(t\right)=\frac{-6\left(t^2+9\right)+12t^2}{\left(t^2+9\right)^2}=\frac{6\left(t^2-9\right)}{\left(t^2+9\right)^2}\)

\(Q'\left(t\right)=0\Rightarrow t=3\)

\(Q\left(0\right)=0\) ; \(Q\left(3\right)=-1\)

Dựa vào BBT, ta thấy \(Q_{min}=-1\) khi \(t=3\Rightarrow x=3\)