Question

2．解方程组$\left\{\begin{array}{l}{{x}^{2}-2xy+{y}^{2}=16}\\{{x}^{2}-9{y}^{2}=0}\end{array}\right.$．

Answer 1

分析 由于组中的两个高次方程都能分解为两个一次方程，所以先分解组中的两个二元二次方程，得到四个一元一次方程，重新组合成二元一次方程组，求出的四个二元一次方程组的解就是原方程的解．

解答 解：$\left\{\begin{array}{l}{{x}^{2}-2xy+{y}^{2}=16①}\\{{x}^{2}-9{y}^{2}=0②}\end{array}\right.$
由①，得（x-y）²=16，
所以x-y=4或x-y=-4．
由②，得（x+3y）（x-3y）=0，
即x+3y=0或x-3y=0
所以原方程组可化为：
$\left\{\begin{array}{l}{x-y=4}\\{x+3y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=4}\\{x-3y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=-4}\\{x+3y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=-4}\\{x-3y=0}\end{array}\right.$
解这些方程组，得
$\left\{\begin{array}{l}{x=3}\\{y=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x=6}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=-3}\\{y=1}\end{array}\right.$，$\left\{\begin{array}{l}{x=-6}\\{y=-2}\end{array}\right.$．
所以原方程组的解为：$\left\{\begin{array}{l}{x=3}\\{y=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x=6}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=-3}\\{y=1}\end{array}\right.$，$\left\{\begin{array}{l}{x=-6}\\{y=-2}\end{array}\right.$．

点评 本题考查了二元二次方程组的解法．解决本题的关键是利用完全平方公式、平方差公式化二元二次方程组为四个一元一次方程组．