Question

Tìm GTLN :

\(M=\dfrac{\sqrt{x}+1}{x\sqrt{x}+7}\)

Answer 1

ĐK: \(x\ge0\)

Mà: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\forall x\left(xđ\right)\\x\sqrt{x}\ge0\forall x\left(xđ\right)\end{matrix}\right.\)

Nên: \(M=\dfrac{\sqrt{x}+1}{x\sqrt{x}+7}\le\dfrac{1}{7}\forall x\left(xđ\right)\)

Dấu "=" xảy ra:

\(\dfrac{\sqrt{x}+1}{x\sqrt{x}+7}=\dfrac{1}{7}\)

\(\Leftrightarrow7\sqrt{x}+7=x\sqrt{x}+7\)

\(\Leftrightarrow x\sqrt{x}-7\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\left(tm\right)\)

Vậy: \(M_{max}=\dfrac{1}{7}\) khi \(\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)