Question

12．解方程組：$\left\{\begin{array}{l}{{x}^{2}-4{y}^{2}=0}\\{{x}^{2}-2xy=8}\end{array}\right.$．

Answer 1

分析 先將①因式分解為：（x-2y）（x+2y）=0，化成兩個(gè)一次方程：x-2y=0和x+2y=0；與②組成兩個(gè)二元二次方程組，解出即可．

解答 解：$\left\{\begin{array}{l}{{x}^{2}-4{y}^{2}=0①}\\{{x}^{2}-2xy=8②}\end{array}\right.$，
由①得 （x-2y）（x+2y）=0③，
由②③得 $\left\{{\begin{array}{l}{x-2y=0}\\{{x^2}-2xy=8}\end{array}}\right.$④$\left\{\begin{array}{l}{x+2y=0}\\{{x}^{2}-2xy=8}\end{array}\right.$ ⑤，
解④得：方程組無(wú)解；
解⑤得：$\left\{{\begin{array}{l}{x=2}\\{y=-1}\end{array}}\right.，\left\{{\begin{array}{l}{x=-2}\\{y=1}\end{array}}\right.$；
∴原方程的解是：$\left\{\begin{array}{l}{{x}_{1}=2}\\{{y}_{1}=-1}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=-2}\\{{y}_{2}=1}\end{array}\right.$．

點(diǎn)評(píng) 本題考查了解高次方程，解高次方程的思路是降次，把它轉(zhuǎn)化成二次方程或一次方程；本題通過(guò)因式分解法達(dá)到降次的目的．