题目内容
9.已知P={6,2x-y,x2-y2},Q={2y,6,x+y},且P=Q,求x,y的值.分析 根据P、Q相等,得到方程组,求出x,y的值,代入集合检验即可.
解答 解:已知P={6,2x-y,x2-y2},Q={2y,6,x+y},且P=Q,
∴$\left\{\begin{array}{l}{2x-y=2y}\\{{x}^{2}{-y}^{2}=x+y}\end{array}\right.$或$\left\{\begin{array}{l}{2x-y=x+y}\\{{x}^{2}{-y}^{2}=2y}\end{array}\right.$,
解得:$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$或$\left\{\begin{array}{l}{x=0}\\{y=0}\end{array}\right.$或$\left\{\begin{array}{l}{x=\frac{4}{3}}\\{y=\frac{2}{3}}\end{array}\right.$,
当x=3,y=2时,P={6,4,5},Q={4,6,5},符合题意;
当x=0,y=0时,P={6,0,0},Q={0,6,0},不合题意;
当x=$\frac{4}{3}$,y=$\frac{2}{3}$时,P={6,2,$\frac{4}{3}$},Q={$\frac{4}{3}$,6,2},符合题意;
综上:$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$或$\left\{\begin{array}{l}{x=\frac{4}{3}}\\{y=\frac{2}{3}}\end{array}\right.$.
点评 本题考查了集合的相等关系,考查集合的含义,是一道基础题.
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