Question

4．解方程组：$\left\{\begin{array}{l}{x}^{2}-3xy+2{y}^{2}=0\\{x}^{2}+{y}^{2}=1\end{array}\right.$．

Answer 1

分析 根据因式分解，可用Y表示x，根据代入消元法，可得方程组的解．

解答 解：$\left\{\begin{array}{l}{{x}^{2}-3xy+2{y}^{2}=0①}\\{{x}^{2}+{y}^{2}=1②}\end{array}\right.$，
由①得（x-y）（x-2y）=0．
于是，得
x-y=0或x-2y=0，
x=y③或x=2y④．
将③代入②得2x²=1，解得x=±$\frac{\sqrt{2}}{2}$，y=$±\frac{\sqrt{2}}{2}$，
方程组的解为$\left\{\begin{array}{l}{x=\frac{\sqrt{2}}{2}}\\{y=\frac{\sqrt{2}}{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\frac{\sqrt{2}}{2}}\\{y=-\frac{\sqrt{2}}{2}}\end{array}\right.$；
将④代入②得5y²=1，解得y=$±\frac{\sqrt{5}}{5}$，x=±$\frac{2\sqrt{5}}{5}$，
方程组的解为$\left\{\begin{array}{l}{x=\frac{2\sqrt{5}}{5}}\\{y=\frac{\sqrt{5}}{5}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\frac{2\sqrt{5}}{5}}\\{y=-\frac{\sqrt{5}}{5}}\end{array}\right.$．
综上所述：原方程组的解为$\left\{\begin{array}{l}{x=\frac{\sqrt{2}}{2}}\\{y=\frac{\sqrt{2}}{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\frac{\sqrt{2}}{2}}\\{y=-\frac{\sqrt{2}}{2}}\end{array}\right.$$\left\{\begin{array}{l}{x=\frac{2\sqrt{5}}{5}}\\{y=\frac{\sqrt{5}}{5}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\frac{2\sqrt{5}}{5}}\\{y=-\frac{\sqrt{5}}{5}}\end{array}\right.$．

点评 本题考查了解方程组，利用因式分解得出x=y或x=2y是解题关键．