Question

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

Answer 1

分析 先把第一个方程利用因式分解的方法化为x-3y=0或x+y=0，则原方程可转化为$\left\{\begin{array}{l}{x-3y=0}\\{{x}^{2}-xy+{y}^{2}=3}\end{array}\right.$或$\left\{\begin{array}{l}{x+y=0}\\{{x}^{2}-xy+{y}^{2}=3}\end{array}\right.$，然后利用代入法解两个二元二次方程组即可．

解答 解：$\left\{\begin{array}{l}{{x}^{2}-2xy-3{y}^{2}=0①}\\{{x}^{2}-xy+{y}^{2}=3②}\end{array}\right.$，
由①得（x-3y）（x+y）=0，
所以x-3y=0或x+y=0，
所以原方程可转化为$\left\{\begin{array}{l}{x-3y=0}\\{{x}^{2}-xy+{y}^{2}=3}\end{array}\right.$或$\left\{\begin{array}{l}{x+y=0}\\{{x}^{2}-xy+{y}^{2}=3}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x=\frac{3\sqrt{21}}{7}}\\{y=\frac{\sqrt{21}}{7}}\end{array}\right.$或$\left\{\begin{array}{l}{x=-\frac{3\sqrt{21}}{7}}\\{y=-\frac{\sqrt{21}}{7}}\end{array}\right.$或$\left\{\begin{array}{l}{x=1}\\{y=-1}\end{array}\right.$或$\left\{\begin{array}{l}{x=-1}\\{y=1}\end{array}\right.$，
所以原方程组的解为$\left\{\begin{array}{l}{x=\frac{3\sqrt{21}}{7}}\\{y=\frac{\sqrt{21}}{7}}\end{array}\right.$或$\left\{\begin{array}{l}{x=-\frac{3\sqrt{21}}{7}}\\{y=-\frac{\sqrt{21}}{7}}\end{array}\right.$或$\left\{\begin{array}{l}{x=1}\\{y=-1}\end{array}\right.$或$\left\{\begin{array}{l}{x=-1}\\{y=1}\end{array}\right.$．

点评 本题考查了高次方程：通过适当的方法，把高次方程化为次数较低的方程求解．所以解高次方程一般要降次，即把它转化成二次方程或一次方程．也有的通过因式分解来解．