题目内容
16.解二元二次方程组$\left\{\begin{array}{l}{5{x}^{2}+2{y}^{2}+x-4y-6=0}\\{2{x}^{2}+{y}^{2}+x-2y-3=0}\end{array}\right.$.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,进而解得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.
∴$\left\{\begin{array}{l}{x=0}\\{y=-1}\end{array}\right.$,$\left\{\begin{array}{l}{x=0}\\{y=3}\end{array}\right.$,$\left\{\begin{array}{l}{x=1}\\{y=0}\end{array}\right.$,$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$.
∴原方程组的解为:$\left\{\begin{array}{l}{x=0}\\{y=-1}\end{array}\right.$,$\left\{\begin{array}{l}{x=0}\\{y=3}\end{array}\right.$,$\left\{\begin{array}{l}{x=1}\\{y=0}\end{array}\right.$,$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$.
点评 本题考查了方程组的解法,考查了推理能力与计算能力,属于中档题.
A. | ① | B. | ①② | C. | ①②③ | D. | ①②③④ |