Question

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

Answer 1

分析 把①化為x=±2y，把②化為x+y=±2，重新組成方程組，解二元一次方程組即可．

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，
則$\left\{\begin{array}{l}{x=2y}\\{x+y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=-2y}\\{x+y=2}\end{array}\right.$，$\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.$．

點(diǎn)評 本題考查的是二元二次方程組的解法，把二元二次方程根據(jù)平方差公式和完全平方公式進(jìn)行變形化為兩個(gè)二元一次方程是解題的關(guān)鍵．