Question

18．解方程组：$\left\{\begin{array}{l}{{x}^{2}-4{y}^{2}=0①}\\{{x}^{2}+2xy+{y}^{2}=4②}\end{array}\right.$．

Answer 1

分析 把①化为x=±2y，把②化为x+y=±2，重新组成方程组，解二元一次方程组即可．

解答 解：$\left\{\begin{array}{l}{{x}^{2}-4{y}^{2}=0①}\\{{x}^{2}+2xy+{y}^{2}=4②}\end{array}\right.$，
由①得，x=±2y，
由②得，x+y=±2，
则$\left\{\begin{array}{l}{x=2y}\\{x+y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=-2y}\\{x+y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=-2y}\\{x+y=-2}\end{array}\right.$$\left\{\begin{array}{l}{x=2y}\\{x+y=-2}\end{array}\right.$，
解得，$\left\{\begin{array}{l}{{x}_{1}=\frac{4}{3}}\\{{y}_{1}=\frac{2}{3}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=4}\\{{y}_{2}=-2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=-4}\\{{y}_{3}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=-\frac{4}{3}}\\{{y}_{4}=-\frac{2}{3}}\end{array}\right.$．

点评 本题考查的是二元二次方程组的解法，把二元二次方程根据平方差公式和完全平方公式进行变形化为两个二元一次方程是解题的关键．