Question

16．解方程：
$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=10}\\{{x}^{2}-4xy+3{y}^{2}=0}\end{array}\right.$．

Answer 1

分析 由②可得（x-y）（x-3y）=0，進而得到2個方程組$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=10}\\{x-y=0}\end{array}\right.$和$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=10}\\{x-3y=0}\end{array}\right.$，再分別解2個方程組求解即可．

解答 解：$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=10①}\\{{x}^{2}-4xy+3{y}^{2}=0②}\end{array}\right.$，
由②可得（x-y）（x-3y）=0，
則$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=10}\\{x-y=0}\end{array}\right.$和$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=10}\\{x-3y=0}\end{array}\right.$，
解得$\left\{\begin{array}{l}{{x}_{1}=-\sqrt{5}}\\{{y}_{1}=-\sqrt{5}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=\sqrt{5}}\\{{y}_{2}=\sqrt{5}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=-3}\\{{y}_{3}=-1}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=3}\\{{y}_{4}=1}\end{array}\right.$．

點評 考查了高次方程，高次方程的解法思想：通過適當?shù)姆椒�，把高次方程化為次�?shù)較低的方程求解．所以解高次方程一般要降次，即把它轉化成二次方程或一次方程．也有的通過因式分解來解．