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\(\alpha\) 為 \(x^2-3x+1=0\) 之一根, \(\beta\) 為 \(x^2-2x-1=0\) 之一根,求

\(\displaystyle\left\{\begin{matrix}\displaystyle\left(\beta^2+\frac{1}{\beta^2}\right)X+\left(\alpha^2+\frac{1}{\alpha^2}\right)Y=\alpha^4+\frac{1}{\alpha^4}  \\ \left(\alpha^3+\frac{1}{\alpha^3}\right)X-\left(\beta^3-\frac{1}{\beta^3}\right)Y=-\left(\beta^4+\frac{1}{\beta^4}\right)\\  \end{matrix} \right.\) 之 \(X\) 解?

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回復 1# thankyou 的帖子

\(\begin{align}
  & {{\alpha }^{2}}-3\alpha +1=0 \\
& {{\alpha }^{2}}+1=3\alpha  \\
& \alpha +\frac{1}{\alpha }=3 \\
&  \\
& \beta -\frac{1}{\beta }=2 \\
&  \\
& {{\alpha }^{2}}+\frac{1}{{{\alpha }^{2}}}=7 \\
& {{\alpha }^{3}}+\frac{1}{{{\alpha }^{3}}}=18 \\
& {{\alpha }^{4}}+\frac{1}{{{\alpha }^{4}}}=47 \\
& {{\beta }^{2}}+\frac{1}{{{\beta }^{2}}}=6 \\
& {{\beta }^{3}}-\frac{1}{{{\beta }^{3}}}=14 \\
& {{\beta }^{4}}+\frac{1}{{{\beta }^{4}}}=34 \\
\end{align}\)

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