Question

[123 nhan]

Tìm GTLN hoặc GTNN của:

\(C=\sqrt{-x^2+6x}\)

\(D=\sqrt{6x-2x^2}\)

Answer 1

[HT.Phong (9A5)]

Ta có: 

\(C=\sqrt{-x^2+6x}\) 

Mà: \(\sqrt{-x^2+6x}\ge0\) 

Dấu "=" xảy ra khi:

\(\sqrt{-x^2+6x}=0\)

\(\Leftrightarrow\sqrt{-x\left(x-6\right)}=0\)

\(\Leftrightarrow-x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: \(C_{min}=0\) khi \(\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Answer 2

[HT.Phong (9A5)]

\(D=\sqrt{6x-2x^2}\)

Mà: \(\sqrt{6x-2x^2}\ge0\)

Dấu "=" xảy ra khi:

\(\sqrt{6x-2x^2}=0\)

\(\Leftrightarrow\sqrt{2x\left(3-x\right)}=0\)

\(\Leftrightarrow2x\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy: \(D_{min}=0\) khi \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)