题目内容
14.设A={(x,y)|2x-y=1},B={(x,y)|5x+y=6},C={(x,y)|2x=y+1},D={(x,y)|2x-y=8}},则A∩B={(1,1)},B∩C={(1,1)},A∩D=∅.分析 直接利用交集运算求解方程组得答案.
解答 解:∵A={(x,y)|2x-y=1},B={(x,y)|5x+y=6},C={(x,y)|2x=y+1},D={(x,y)|2x-y=8}},
∴A∩B={(x,y)|$\left\{\begin{array}{l}{2x-y=1}\\{5x+y=6}\end{array}\right.$}={(1,1)},
B∩C={(x,y)|$\left\{\begin{array}{l}{5x+y=6}\\{2x=y+1}\end{array}\right.$}={(1,1)},
A∩D={(x,y)|$\left\{\begin{array}{l}{2x-y=1}\\{2x-y=8}\end{array}\right.$}=∅.
故答案为:{(1,1)},{(1,1)},∅.
点评 本题考查交集及其运算,考查了方程组的解法,是基础题.
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