上中自招面试数学

$1$

方法一

\[\begin{align*} &x^2 - 3x + 1 = 0\\ &x^2 = 3x - 1,\ \frac{1}{x} = 3 - x\\ & \alpha + \beta = 3,\ \alpha\beta=1\\ &(\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta=5\\ &\beta-\alpha=\sqrt5\\ &\begin{aligned} \frac{2}{\alpha} + \beta^3 &= \frac{1+\alpha\beta^3}{\alpha}\\&= \frac{1+3\beta}{\alpha}\\&=\frac{1+3\beta+3\alpha-3\alpha}{\alpha}\\&=\frac{10}{\alpha}-3\\&=10\beta-3\\&=5\beta+5(3-\alpha)-3\\&=5(\beta-\alpha)+12\\&=12+5\sqrt5 &\end{aligned} \end{align*}\]

方法二

\[\begin{align*} &\frac{2}{\alpha} + \beta^3 = 6 - 2\alpha + 3\beta^2 - \beta = 3 + 8\beta - 2\alpha\\ &\text{设} \ a = 8\beta - 2\alpha,\ b = 8\alpha - 2\beta\\ &a + b = 6(\alpha + \beta) = 18\\ &a- b = 10(\beta-\alpha)=10\sqrt{(\alpha+\beta)^2-4\alpha\beta}=10\sqrt5\\ &a = 9 + 5\sqrt{5}\\ &\text{原式} = 3 + a = 12+5\sqrt5 \end{align*}\]

$2$

当 $ab+\stackrel\frown{acb} =k(ab+\stackrel\frown{ab})$，k为大于1的整数时，两只小虫可不相遇。

不一定对，还需要进一步的讨论