Question

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

Answer 1

分析 由x²-3xy+2y²=0可得x=y或x=2y．分別代入x²+y²=10解出即可．

解答 解：$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=10}&{①}\\{{x}^{2}-3xy+{2y}^{2}=0}&{②}\end{array}\right.$，
由②可得x=y或x=2y．
把x=y代入①可得y²=5，解得y=$±\sqrt{5}$．
∴$\left\{\begin{array}{l}{x=\sqrt{5}}\\{y=\sqrt{5}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\sqrt{5}}\\{y=-\sqrt{5}}\end{array}\right.$．
把x=2y代入①可得y²=2，解得y=±$\sqrt{2}$．
∴$\left\{\begin{array}{l}{x=2\sqrt{2}}\\{y=\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-2\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$．
綜上可得原方程組的解為：$\left\{\begin{array}{l}{x=\sqrt{5}}\\{y=\sqrt{5}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\sqrt{5}}\\{y=-\sqrt{5}}\end{array}\right.$，$\left\{\begin{array}{l}{x=2\sqrt{2}}\\{y=\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-2\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$．

點(diǎn)評(píng) 本題考查了“代入消元法”解方程組，考查了計(jì)算能力，屬于中檔題．