Question

20．解方程组：$\left\{\begin{array}{l}{{x}^{2}-5xy+6{y}^{2}=0①}\\{{x}^{2}+{y}^{2}=20②}\end{array}\right.$．

Answer 1

分析 首先对原方程组进行化简，用含y的表达式表示出x，然后分别重新组合，成为两个方程组，最后解这两个方程组即可．

解答 解：原方程组可化为如下两个方程组
（1）$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=20}\\{x=2y}\end{array}\right.$（2）$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=20}\\{x=3y}\end{array}\right.$
解方程组（1）得$\left\{\begin{array}{l}{{x}_{1}=4}\\{{y}_{1}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=-4}\\{{y}_{2}=-2}\end{array}\right.$
解方程组（2）得$\left\{\begin{array}{l}{{x}_{3}=3\sqrt{2}}\\{{y}_{3}=\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=-3\sqrt{2}}\\{{y}_{4}=-\sqrt{2}}\end{array}\right.$
∴原方程组的解为$\left\{\begin{array}{l}{{x}_{1}=4}\\{{y}_{1}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=-4}\\{{y}_{2}=-2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=3\sqrt{2}}\\{{y}_{3}=\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=-3\sqrt{2}}\\{{y}_{4}=-\sqrt{2}}\end{array}\right.$

点评 本题主要考查解高次方程，关键在于对原方程组的两个方程进行化简，重新组合．