Question

15．解方程组：$\left\{{\begin{array}{l}{{x^2}-{y^2}=1}\\{{x^2}-3xy-4{y^2}=0}\end{array}}\right.$．

Answer 1

分析 先把第二个方程因式分解，把二元二次方程组转化为二元一次方程组，求解即可．

解答 解：由②得 x-4y=0或x+3y=0，
原方程组可化为（Ⅰ）$\left\{\begin{array}{l}{x^2}-{y^2}=1\\ x-4y=0\end{array}\right.$（Ⅱ）$\left\{\begin{array}{l}{x^2}-{y^2}=1\\ x+y=0\end{array}\right.$，
解方程组（Ⅰ）得$\left\{\begin{array}{l}{x_1}=\frac{{4\sqrt{15}}}{15}\\{y_1}=\frac{{\sqrt{15}}}{15}\end{array}\right.\left\{\begin{array}{l}{x_2}=-\frac{{4\sqrt{15}}}{15}\\{y_2}=-\frac{{\sqrt{15}}}{15}\end{array}\right.$，
方程组（Ⅱ）无解，
所以原方程组的解是$\left\{\begin{array}{l}{x_1}=\frac{{4\sqrt{15}}}{15}\\{y_1}=\frac{{\sqrt{15}}}{15}\end{array}\right.\left\{\begin{array}{l}{x_2}=-\frac{{4\sqrt{15}}}{15}\\{y_2}=-\frac{{\sqrt{15}}}{15}\end{array}\right.$．

点评 本题考查了高次方程的解法，解方程组的思想是把二元二次方程组转化为二元一次方程组．