Lời giải:
ĐK: \(x\geq 1; y\geq 4\)
Áp dụng BĐT AM-GM:
\(\sqrt{x-1}=\sqrt{1(x-1)}\leq \frac{x-1+1}{2}=\frac{x}{y}\)
\(\Rightarrow y\sqrt{x-1}\leq \frac{xy}{2}\)
\(\sqrt{y-4}=\frac{1}{2}\sqrt{4(y-4)}\leq \frac{4+(y-4)}{4}=\frac{y}{4}\)
\(\Rightarrow x\sqrt{y-4}\leq \frac{xy}{4}\)
Do đó: \(M\leq \frac{\frac{xy}{2}+\frac{xy}{4}}{xy}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
Vậy \(M_{\max}=\frac{3}{4}\Leftrightarrow x=2; y=8\)
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