1߳\(\sqrt{x}\) \( \;\; \frac{x-y}{\sqrt{xy}+y}=1 \)
³y\(u=x/y\),ԭ̱Ϊ: \(\;\; \frac{u-1}{\sqrt{u}+1}=1\)
ĸϼΪ\(\;\; \sqrt{u}-1=1\), \(x/y=4\)Ϊ2.
2Ҫ˳ָ
\( f(x)=(1-\frac{1}{x})^x\)ǵ\(e^{-1}\)Ϊޡü
\( f(4)=81/256>0.316\) \(.316^2>.099 \)
\[\begin{align}
&&(95/99)^{97}\times 95\\
&=& 1-4/99)^{97}\times 95 \\
&= &(1-4/99)^{99/4\times 4}\times (99^2/95) \\
&\geq& .099^2 (99)^2 /95 \\
&>& 9.8^2 /95 \\
&>&1
\end{align}\]
\( 95^{98}>99^{97}\)
³y\(u=x/y\),ԭ̱Ϊ: \(\;\; \frac{u-1}{\sqrt{u}+1}=1\)
ĸϼΪ\(\;\; \sqrt{u}-1=1\), \(x/y=4\)Ϊ2.
2Ҫ˳ָ
\( f(x)=(1-\frac{1}{x})^x\)ǵ\(e^{-1}\)Ϊޡü
\( f(4)=81/256>0.316\) \(.316^2>.099 \)
\[\begin{align}
&&(95/99)^{97}\times 95\\
&=& 1-4/99)^{97}\times 95 \\
&= &(1-4/99)^{99/4\times 4}\times (99^2/95) \\
&\geq& .099^2 (99)^2 /95 \\
&>& 9.8^2 /95 \\
&>&1
\end{align}\]
\( 95^{98}>99^{97}\)
���༭ʱ��: 2021-09-26 15:26:51



