高一数学必修一集合 奥赛题

\u9ad8\u4e00\u6570\u5b66\u5fc5\u4fee\u4e00\u96c6\u5408\u5965\u8d5b\u9898

{A-2}={0,2,4,6,7} {B+2}={3,4,5,7,10} \u6240\u4ee5{A-2}={B+2}={4,7}C={4}\u6216{7},{4, 7}

\u53bb http://wenku.baidu.com/view/fa764645a8956bec0975e392.html\u4e0b\u8f7d\uff01\u90e8\u5206\u5185\u5bb9\u5982\u4e0b\uff1a
\u96c6\u5408\uff0824\u4e4b\u540e\u6709\u7b54\u9898\uff09
1. \u8bbe\u96c6\u5408A={-4\uff0c2m-1,m2}\uff0cB={9\uff0cm-5\uff0c1-m}\uff0c\u53c8AB={9},\u6c42\u5b9e\u6570m\u7684\u503c.
2. \u8bbeA={x|x2+ax+b=0},B={x|x2+cx+15=0},\u53c8AB={3\uff0c5}\uff0cA\u2229B={3}\uff0c\u6c42\u5b9e\u6570a,b,c\u7684\u503c.
3. \u8bbeU=\uff5b1,2,3,4,5,6,7,8\uff5d,A=\uff5b3,4,5\uff5d,B=\uff5b4,7,8\uff5d,\u6c42CuA,
CuB, (CuA) (CuB),
(CuA) (CuB),
Cu(AB) , Cu(AB)\uff0e
4. \u5df2\u77e5\u5168\u96c6U={x|x2-3x+2\u22650}\uff0cA={x||x-2|>1}\uff0cB=\uff0c
\u6c42CUA\uff0cCUB\uff0cA\u2229B\uff0cA\u2229(CUB)\uff0c(CUA)\u2229B
\u2026\u2026
23. \u5df2\u77e5\uff1a\u96c6\u5408A={x|\u22640}\uff0c B={x|x2\uff0d3x+2<0}\uff0cU=R\uff0c
\u6c42\uff081\uff09A\u222aB\uff1b
\uff082\uff09\uff08uA\uff09\u2229B.
24. \u96c6\u5408A\uff1d\uff5bx\uff5cx 2\uff0dax\uff0ba2\uff0d19\uff1d0\uff5d\uff0cB\uff1d\uff5bx\uff5cx 2\uff0d5 x\uff0b6\uff1d0\uff5d\uff0cC\uff1d\uff5bx\uff5cx 2\uff0b2 x\uff0d8\uff1d0\uff5d\uff0e
\uff081\uff09\u82e5A\u2229B\uff1dA\u222aB\uff0c\u6c42a\u7684\u503c\uff1b\uff082\uff09\u82e5 A\u2229B\uff0cA\u2229C\uff1d\uff0c\u6c42a\u7684\u503c\uff0e

\u3010\u3010\u4e0d\u6e05\u695a\uff0c\u518d\u95ee\uff1b\u6ee1\u610f\uff0c \u8bf7\u91c7\u7eb3\uff01\u795d\u4f60\u597d\u8fd0\u5f00\u2606\uff01\uff01\u3011\u3011

解：由题意得:C+2包含于A,所以C包含于A-2,即:C包含于{0、2、4、6、7}
同理C-2包含于B,所以C包含于B+2,即:C包含于{3、4、5、7、10}
所以C同时包含于{4、7}
所以C应该为{4}，{7},或{4，7}
希望能够帮你!

{A-2}={0,2,4,6,7}
{B+2}={3,4,5,7,10}
所以{A-2}={B+2}={4,7}C={4}或{7},{4,<br/>7}

｛4,7}｛4}｛7｝

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