Question

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

Answer 1

分析 根據因式分解，可用Y表示x，根據代入消元法，可得方程組的解．

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．
將③代入②得2x²=1，解得x=±$\frac{\sqrt{2}}{2}$，y=$±\frac{\sqrt{2}}{2}$，
方程組的解為$\left\{\begin{array}{l}{x=\frac{\sqrt{2}}{2}}\\{y=\frac{\sqrt{2}}{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\frac{\sqrt{2}}{2}}\\{y=-\frac{\sqrt{2}}{2}}\end{array}\right.$；
將④代入②得5y²=1，解得y=$±\frac{\sqrt{5}}{5}$，x=±$\frac{2\sqrt{5}}{5}$，
方程組的解為$\left\{\begin{array}{l}{x=\frac{2\sqrt{5}}{5}}\\{y=\frac{\sqrt{5}}{5}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\frac{2\sqrt{5}}{5}}\\{y=-\frac{\sqrt{5}}{5}}\end{array}\right.$．
綜上所述：原方程組的解為$\left\{\begin{array}{l}{x=\frac{\sqrt{2}}{2}}\\{y=\frac{\sqrt{2}}{2}}\end{array}\right.$，$\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.$．

點評 本題考查了解方程組，利用因式分解得出x=y或x=2y是解題關鍵．