Question

16．解二元二次方程組$\left\{\begin{array}{l}{5{x}^{2}+2{y}^{2}+x-4y-6=0}\\{2{x}^{2}+{y}^{2}+x-2y-3=0}\end{array}\right.$．

Answer 1

分析 $\left\{\begin{array}{l}{5{x}^{2}+2{y}^{2}+x-4y-6=0}&{①}\\{2{x}^{2}+{y}^{2}+x-2y-3=0}&{②}\end{array}\right.$，①-②×2可得：x²-x=0，解得x，進(jìn)而解得y．

解答 解：$\left\{\begin{array}{l}{5{x}^{2}+2{y}^{2}+x-4y-6=0}&{①}\\{2{x}^{2}+{y}^{2}+x-2y-3=0}&{②}\end{array}\right.$，
①-②×2可得：x²-x=0，解得x=0或1．
∴$\left\{\begin{array}{l}{x=0}\\{y=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x=0}\\{y=3}\end{array}\right.$，$\left\{\begin{array}{l}{x=1}\\{y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$．
∴原方程組的解為：$\left\{\begin{array}{l}{x=0}\\{y=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x=0}\\{y=3}\end{array}\right.$，$\left\{\begin{array}{l}{x=1}\\{y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$．

點(diǎn)評(píng) 本題考查了方程組的解法，考查了推理能力與計(jì)算能力，屬于中檔題．