Question

10．解方程組：$\left\{\begin{array}{l}x+3y=6\\{x^2}-4xy+4{y^2}=1.\end{array}\right.$．

Answer 1

分析 由方程②可得（x-2y）²=1即x-2y=1或x-2y=-1，將原方程組分為$\left\{\begin{array}{l}x+3y=6\\ x-2y=1\end{array}\right.$或$\left\{\begin{array}{l}x+3y=6\\ x-2y=-1\end{array}\right.$，分別解每個方程組可得．

解答 解：在方程組$\left\{\begin{array}{l}{x+3y=6}&{①}\\{{x}^{2}-4xy+4{y}^{2}=1}&{②}\end{array}\right.$中，
由②得：（x-2y）²=1，
∴x-2y=1或x-2y=-1，
所以原方程組變?yōu)椋?\left\{\begin{array}{l}x+3y=6\\ x-2y=1\end{array}\right.$或$\left\{\begin{array}{l}x+3y=6\\ x-2y=-1\end{array}\right.$，
解這兩個方程組得：$\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$，$\left\{\begin{array}{l}x=\frac{9}{5}\\ y=\frac{7}{5}\end{array}\right.$
所以原方程組的解為$\left\{\begin{array}{l}{x_1}=3\\{y_2}=1\end{array}\right.$，$\left\{\begin{array}{l}{x_2}=\frac{9}{5}\\{y_2}=\frac{7}{5}\end{array}\right.$．

點評 本題主要考查解高次方程組的能力，體現(xiàn)了化歸思想在解高次方程或多元方程中的應(yīng)用，解高次方程需降冪，多元方程需消元．