Question

20．解方程組：$\left\{\begin{array}{l}{{x}^{2}-5xy+6{y}^{2}=0①}\\{{x}^{2}+{y}^{2}=20②}\end{array}\right.$．

Answer 1

分析 首先對原方程組進(jìn)行化簡，用含y的表達(dá)式表示出x，然后分別重新組合，成為兩個方程組，最后解這兩個方程組即可．

解答 解：原方程組可化為如下兩個方程組
（1）$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=20}\\{x=2y}\end{array}\right.$（2）$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=20}\\{x=3y}\end{array}\right.$
解方程組（1）得$\left\{\begin{array}{l}{{x}_{1}=4}\\{{y}_{1}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=-4}\\{{y}_{2}=-2}\end{array}\right.$
解方程組（2）得$\left\{\begin{array}{l}{{x}_{3}=3\sqrt{2}}\\{{y}_{3}=\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=-3\sqrt{2}}\\{{y}_{4}=-\sqrt{2}}\end{array}\right.$
∴原方程組的解為$\left\{\begin{array}{l}{{x}_{1}=4}\\{{y}_{1}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=-4}\\{{y}_{2}=-2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=3\sqrt{2}}\\{{y}_{3}=\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=-3\sqrt{2}}\\{{y}_{4}=-\sqrt{2}}\end{array}\right.$

點(diǎn)評 本題主要考查解高次方程，關(guān)鍵在于對原方程組的兩個方程進(jìn)行化簡，重新組合．