Question

14．若方程组$\left\{\begin{array}{l}{ax-2y=1}\\{ax+2y=5}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=a}\\{y=b}\end{array}\right.$，则$\frac{a}{b}$=±$\sqrt{3}$．

Answer 1

分析 根据题意得出$\left\{\begin{array}{l}{{a}^{2}-2b=1}\\{{a}^{2}+2b=5}\end{array}\right.$，求出a，b的值，再代入要求的式子即可得出答案．

解答 解：∵方程组$\left\{\begin{array}{l}{ax-2y=1}\\{ax+2y=5}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=a}\\{y=b}\end{array}\right.$，
∴$\left\{\begin{array}{l}{{a}^{2}-2b=1}\\{{a}^{2}+2b=5}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=\sqrt{3}}\\{b=1}\end{array}\right.$或$\left\{\begin{array}{l}{a=-\sqrt{3}}\\{b=1}\end{array}\right.$，
∴$\frac{a}{b}$=$\frac{\sqrt{3}}{1}$=$\sqrt{3}$或$\frac{a}{b}$=$\frac{-\sqrt{3}}{1}$=-$\sqrt{3}$；
故答案为：±$\sqrt{3}$．

点评 本题考查了二元一次方程组的解：一般地二元一次方程组的两个方程的公共解，叫做二元一次方程组的解．