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青青子衿
发表于 2023-7-6 13:32
本帖最后由 青青子衿 于 2023-7-27 10:02 编辑
\begin{align*}
\left(\frac{\left(1-\sqrt{3}\right) u^2+2 \left(1+\sqrt{1+u^3}\right)}
{\left(1+\sqrt{3}\right) u^2+2 \left(1+\sqrt{1+u^3}\right)}\right)^2
+\sqrt{3}\left(\frac{2 u \left(\sqrt{1+u}+\sqrt{1-u+u^2}\right)}{\left(1+\sqrt{3}\right) u^2+2 \left(1+\sqrt{1+u^3}\right)}\right)^2&=1\\
\left(\frac{(2-u) \sqrt{1+u}+(2+u) \sqrt{1-u+u^2}}
{\left(1+\sqrt{3}\right) u^2+2 \left(1+\sqrt{1+u^3}\right)}\right)^2
+\frac{3+2\sqrt{3}}{4}\left(\frac{2 u \left(\sqrt{1+u}+\sqrt{1-u+u^2}\right)}{\left(1+\sqrt{3}\right) u^2+2 \left(1+\sqrt{1+u^3}\right)}\right)^2&=1
\end{align*}
\begin{align*}
&\quad\left(\frac{\left(256+256 x^3+960 x^6+232 x^9+x^{12}\right)-32 \left(8-12 x^3-21 x^6-x^9\right) \sqrt{1+x^3}}{x^3 \left(8-x^3\right)^3}\right)^2\\
&=1+\left(\frac{32 \left(1+x^3\right)^2-4 \left(8-20 x^3-x^6\right) \sqrt{1+x^3}}{x^2 \left(8-x^3\right)^2}\right)^3
\end{align*}
- \left(\frac{32\left(1+t^{3}\right)^{2}-4\left(8-20t^{3}-t^{6}\right)\sqrt{1+t^{3}}}{t^{2}\left(8-t^{3}\right)^{2}},\frac{\left(256+256t^{3}+960t^{6}+232t^{9}+t^{12}\right)-32\left(8-12t^{3}-21t^{6}-t^{9}\right)\sqrt{1+t^{3}}}{t^{3}\left(8-t^{3}\right)^{3}}\right)
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