高一数学 必修一 集合

\u9ad8\u4e00\u6570\u5b66\u5fc5\u4fee1\u96c6\u5408\u95ee\u9898

\u5982\u679cA\u96c6\u5408\u4e2d\u7684\u5143\u7d20\u5728B\u96c6\u5408\u4e2d\u90fd\u5b58\u5728\uff0c\u5219A\u662fB\u7684\u5b50\u96c6

\u5b50\u96c6\u53ea\u5173\u6ce8\u96c6\u5408\u4e2d\u5143\u7d20\u7684\u5b58\u5728\u6027\u3002

A={x\u4e28x\u83f1\u5f62}\uff0cB= {x\u4e28x\u5e73\u884c\u56db\u8fb9\u5f62}

\u5728A\u4e2d\u4efb\u53d6\u83f1\u5f62ABCD\uff0c\u7531\u4e8e\u83f1\u5f62ABCD\u4e5f\u6ee1\u8db3\u5e73\u884c\u56db\u8fb9\u5f62\u7684\u5b9a\u4e49\uff0c\u6240\u4ee5\u83f1\u5f62ABCD\u4e5f\u662fB\u96c6\u5408\u7684\u5143\u7d20\u3002\u6240\u4ee5A\u662fB\u7684\u5b50\u96c6
\u5728B\u96c6\u5408\u4e2d\u4efb\u53d6\u5e73\u884c\u56db\u8fb9\u5f62MNPQ\uff0c\u5982\u679c\u5b83\u4e0d\u5177\u6709\u83f1\u5f62\u7684\u6027\u8d28\uff0c\u5219\u5e73\u884c\u56db\u8fb9\u5f62MNPQ\u4e0d\u662fA\u96c6\u5408\u4e2d\u7684\u5143\u7d20\u3002\u56e0\u6b64\uff0cB\u96c6\u5408\u4e2d\u6709\u7740A\u96c6\u5408\u91cc\u4e0d\u5b58\u5728\u7684\u5143\u7d20\uff0cB\u4e0d\u662fA\u7684\u5b50\u96c6\u3002
\u6240\u4ee5A\u548cB\u4e0d\u76f8\u7b49\uff0cA\u662fB\u7684\u771f\u5b50\u96c6

\u7b2c\u4e8c\u9898\u540c\u7406\u3002

\u5df2\u6bd5\u4e1a\u7684\u9ad8\u4e09\u5b66\u59d0\u7b54\uff1aAUB\u5c31\u662fA\u5e76B\u7684\u610f\u601d\uff0c\u5982\u679cA={1\uff0c2\uff0c3}\uff0cB={4}\uff0c\u90a3\u4e48AUB={1\uff0c2\uff0c3\uff0c4}\u5b50\u96c6\u548c\u771f\u5b50\u96c6\u6700\u5927\u7684\u533a\u522b\u5c31\u662f\uff1a\u5b50\u96c6\u53ef\u4ee5\u5305\u62ec\u81ea\u8eab\uff0c\u800c\u771f\u5b50\u96c6\u4e0d\u53ef\u4ee5\u3002\u901a\u4fd7\u70b9\u8bb2\uff0cA={1\uff0c2\uff0c3}\uff0c\u90a3\u4e48\uff0cA\u7684\u5b50\u96c6\u4e3a{1}\uff0c{2}\uff0c{3}\uff0c{1\uff0c2}\uff0c{1\uff0c3}\uff0c{2\uff0c3}\uff0c{1\uff0c2\uff0c3}\uff0c\u800cA\u7684\u771f\u5b50\u96c6\u4e3a{1}\uff0c{2}\uff0c{3}\uff0c{1\uff0c2}\uff0c{1\uff0c3}\uff0c{2\uff0c3}\u3002\u5173\u4e8e+\uff0c\u6211\u4e0d\u660e\u767d\u4f60\u95ee\u4ec0\u4e48\u3002\u8fd8\u6709\u4e0d\u660e\u767d\u7684\u6b22\u8fce\u8ffd\u95ee\u3002

|x-1|＞2
x-1＞2或x-1＜-2
解得x＞3或x＜-1
即A={x|x＞3或x＜-1}
CuA={x|-1≤x≤3}
x^2-6x+8
=(x-2)(x-4)＜0
∴2＜x＜4
∴B={x|2＜x＜4}
CuA∩B
={x|-1≤x≤3}∩{x|2＜x＜4}
={x|2＜x≤3}

|x-1|>2<=>x-1>2或x-1<-2
x^2-6x+8<0<=>(x-2)(x-4)<0好了剩下的自己做

A={x|x<-1或x>3}
B={x|2<x<4}
CuA={-1<=x<=3}
所以答案是=｛x|2<x<=3}

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