Question

10．解方程组$\left\{\begin{array}{l}{{x}^{2}+4{y}^{2}=4}\\{\sqrt{3}x+2y=2}\end{array}\right.$．

Answer 1

分析 由②得$4{y}^{2}=4-4\sqrt{3}x+3{x}^{2}$③，把③代入①解答即可．

解答 解：$\left\{\begin{array}{l}{{x}^{2}+4{y}^{2}=4①}\\{\sqrt{3}x+2y=2②}\end{array}\right.$，由②得$4{y}^{2}=4-4\sqrt{3}x+3{x}^{2}$③，
把③代入①得：${x}^{2}-\sqrt{3}x=0$，
解得：${x}_{1}=0，{x}_{2}=\sqrt{3}$，
当x₁=0时，y₁=1；
当${x}_{2}=\sqrt{3}$时，${y}_{2}=-\frac{1}{2}$，
所以方程组的解是$\left\{\begin{array}{l}{{x}_{1}=0}\\{{y}_{1}=1}\end{array}\right.，\left\{\begin{array}{l}{{x}_{2}=\sqrt{3}}\\{{y}_{2}=-\frac{1}{2}}\end{array}\right.$．

点评 此题考查高次方程问题，关键是把高次方程化为一般方程再解答．