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請教台南一中科學班98複試2題、103初試4題

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回復 1# ycdye 的帖子

"一個圓的兩條弦"那題,題目有問題,BE不唯一

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回復 1# ycdye 的帖子

2
\(\begin{align}
  & {{T}_{n}}=\frac{2}{3}\times \frac{4}{3}\times \frac{4}{5}\times \frac{6}{5}\times \cdots \cdots \times \frac{2n}{2n+1}\times \frac{2n+2}{2n+1} \\
& {{T}_{n}}^{2}>\frac{1}{2}\times \frac{3}{2}\times \frac{2}{3}\times \frac{4}{3}\times \frac{3}{4}\times \frac{5}{4}\times \cdots \cdots \times \frac{2n-1}{2n}\times \frac{2n+1}{2n}\times \frac{2n}{2n+1}\times \frac{2n+2}{2n+1}=\frac{n+1}{2n+1} \\
& {{T}_{n}}^{2}<\frac{2}{3}\times \frac{4}{3}\times \frac{3}{4}\times \frac{5}{4}\times \frac{4}{5}\times \frac{6}{5}\times \cdots \cdots \times \frac{2n}{2n+1}\times \frac{2n+2}{2n+1}\times \frac{2n+1}{2n+2}\times \frac{2n+3}{2n+2}=\frac{2n+3}{3n+3} \\
\end{align}\)

2
\(\begin{align}
  & {{x}^{2}}+2xy-1=0 \\
& 2{{y}^{2}}+xy-2=0 \\
& 2\left( {{x}^{2}}+2xy \right)=2{{y}^{2}}+xy=2 \\
& 2{{x}^{2}}+3xy-2{{y}^{2}}=0 \\
& \left( x+2y \right)\left( 2x-y \right)=0 \\
& y=2x \\
& ...... \\
&  \\
\end{align}\)

2
\(\begin{align}
  & {{x}^{2}}+xy+{{y}^{2}}-1=0 \\
& {{y}^{2}}-4\left( {{y}^{2}}-1 \right)\ge 0 \\
& ...... \\
\end{align}\)

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回復 1# ycdye 的帖子

15
\(\begin{align}
  & x\left( {{x}^{2}}-yz \right)=3 \\
& y\left( {{y}^{2}}-xz \right)=9 \\
& z\left( {{z}^{2}}-xy \right)=30 \\
&  \\
& xyz={{x}^{3}}-3={{y}^{3}}-9={{z}^{3}}-30 \\
& y=\sqrt[3]{{{x}^{3}}+6},z=\sqrt[3]{{{x}^{3}}+27} \\
&  \\
& x\sqrt[3]{{{x}^{3}}+6}\sqrt[3]{{{x}^{3}}+27}={{x}^{3}}-3 \\
& {{x}^{3}}\left( {{x}^{3}}+6 \right)\left( {{x}^{3}}+27 \right)={{\left( {{x}^{3}}-3 \right)}^{3}} \\
& 42{{x}^{6}}+135{{x}^{3}}+27=0 \\
& \left( {{x}^{3}}+3 \right)\left( 42{{x}^{3}}+9 \right)=0 \\
& {{x}^{3}}=-3\ or\ -\frac{3}{14} \\
& xyz=-6\ or\ -\frac{45}{14} \\
& \left( x,y,z \right)=\left( -\sqrt[3]{3},\sqrt[3]{3},2\sqrt[3]{3} \right) \\
\end{align}\)

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回復 10# ycdye 的帖子

∠BAC 用銳角和鈍角分別畫圖,就知道 BE 不唯一了

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