Question

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

Answer 1

分析 先将原方程组进行变形，利用代入法和换元法可以解答本题．

解答 解：$\left\{\begin{array}{l}{xy=3}&{①}\\{{x}^{2}-2xy+{y}^{2}=4}&{②}\end{array}\right.$，
由①，得
$y=\frac{3}{x}$③，
将①③代入②，得
${x}^{2}-6+\frac{9}{{x}^{2}}=4$，
设x²=t，
则$t-6+\frac{9}{t}=4$，
即t²-10t+9=0，
解得，t=1或t=9，
∴x²=1或x²=9，
解得x=±1或x=±3，
则$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$或$\left\{\begin{array}{l}{x=-1}\\{y=-3}\end{array}\right.$或$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=-3}\\{y=-1}\end{array}\right.$，
即原方程组的解是：$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$或$\left\{\begin{array}{l}{x=-1}\\{y=-3}\end{array}\right.$或$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=-3}\\{y=-1}\end{array}\right.$．

点评 本题考查高次方程，解题的关键是明确解高次方程的方法，尤其是注意换元法的应用．