12道高中数学难题!十一放假前求解 高中数学题!求解!在线等!急

\u9ad8\u4e00\u6570\u5b66\u96be\u9898\uff0c\u6025\u6c42\u89e3\uff01\uff01\uff01

\u4e00\u3001\u57fa\u7840\u77e5\u8bc6\uff1a(\u53c2\u9605\u300a\u91d1\u724c\u4e4b\u8def•\u7ade\u8d5b\u8f85\u5bfc•\u9ad8\u4e2d\u6570\u5b66\u300b\u7b2c\u4e00\u8bb2\uff1a\u96c6\u5408\uff1b\u7b2c\u4e09\u5341\u516b\u8bb2\uff1a\u5bb9\u65a5\u539f\u7406\uff1b\u300a\u91d1\u724c\u4e4b\u8def•\u7ade\u8d5b\u89e3\u9898\u6307\u5bfc•\u9ad8\u4e2d\u6570\u5b66\u300b\u7b2c2\u8bb2\uff1a\u96c6\u5408)
1. \u5143\u7d20\u4e0e\u96c6\u5408\uff1aa\u2208A,bA
2. \u96c6\u5408\u4e0e\u96c6\u5408\uff1aA B,AB,AB,A\u2229B,A\u222aB, UA,\u2026\u2026
3. \u5dee\u96c6\uff1aA\uff0dB\uff1d{x|x\u2208A\u4e14xB}(\u90e8\u5206\u8d44\u6599\u4e0a\u7528\u201cA\B\u201d\u8868\u793a)
4. \u96c6\u5408\u8fd0\u7b97\u5f8b\uff1a(\u7565)
5. n\u4e2a\u5143\u7d20\u7684\u96c6\u5408\u6240\u6709\u5b50\u96c6\u4e2a\u6570\u4e3a\uff1a2n
6. \u8986\u76d6\u4e0e\u5212\u5206\uff1a\u5982\u679c\u96c6\u5408S\uff1dS1\u222aS2\u222a\u2026\u2026\u222aSn\uff0c\u5219S1\u3001S2\u3001\u2026\u2026\u3001Sn\u53eb\u505a\u96c6\u5408S\u7684\u4e00\u4e2a\u8986\u76d6\uff1b\u5982\u679c\u540c\u65f6\u53c8\u6709Si\u2229Sj\uff1d\u03c6(i\u2260j),\u5219S1\u3001S2\u3001\u2026\u2026\u3001Sn\u53eb\u505a\u96c6\u5408S\u7684\u4e00\u4e2a\u5212\u5206.
7. \u5bb9\u65a5\u539f\u7406\uff1acard(A\u222aB)\uff1dcard(A)\uff0bcard(B)\uff0dcard(A\u2229B)
card(A\u222aB\u222aC)\uff1dcard(A)\uff0bcard(B)\uff0bcard(C)
\uff0dcard(A\u2229B)\uff0dcard(B\u2229C)\uff0dcard(C\u2229A)
\uff0bcard(A\u2229B\u2229C)
\u8be5\u7ed3\u8bba\u53ef\u4ee5\u63a8\u5e7f\u5230n\u4e2a\u96c6\u5408.
8. \u547d\u9898\u4e0e\u63a8\u7406\uff1a\u7b80\u5355\u547d\u9898\u4e0e\u590d\u5408\u547d\u9898\uff0c\u903b\u8f91\u5173\u8fde\u8bcd\u201c\u6216\u201d\u3001\u201c\u4e14\u201d\u3001\u201c\u975e\u201d\u7684\u5e94\u7528\uff0c\u9006\u547d\u9898\u3001\u5426\u547d\u9898\u3001\u9006\u5426\u547d\u9898\u53ca\u5176\u771f\u5047\u6027\u7684\u5224\u65ad
9. \u5145\u8981\u6761\u4ef6\uff1a\u5982\u679cAB\uff0c\u5219\u79f0A\u662fB\u7684\u5145\u5206\u6761\u4ef6\uff0c\u540c\u65f6\u79f0B\u662fA\u7684\u5fc5\u8981\u6761\u4ef6
10. \u6570\u5b66\u6096\u8bba\uff1a\u5bf9\u4e8e\u547d\u9898p\uff0c\u5982\u679cp\u6b63\u786e\uff0c\u5219\u53ef\u4ee5\u63a8\u5bfc\u51fa\u201c\u975ep\u201d\uff0c\u800c\u5982\u679cp\u9519\u8bef\uff0c\u53c8\u53ef\u4ee5\u63a8\u5bfc\u51fap\u6b63\u786e\u3002\u4e5f\u79f0\u201c\u4e8c\u96be\u95ee\u9898\u201d\u3002
\u4e8c\u3001\u4f8b\u9898\uff1a
1. \u5df2\u77e5\u96c6\u5408A\uff1d{1\uff0c3\uff0cx}\uff0cB\uff1d{1\uff0cx2}\uff0cA\u222aB\uff1d{1\uff0c3\uff0cx}\uff0c\u5219\u8fd9\u6837\u7684x\u7684\u4e0d\u540c\u7684\u503c\u6709( )\u4e2a
A.1 B.2 C.3 D.4\u5df2\u77e5\u96c6\u5408M\u4e2d\u7684\u5143\u7d20\u90fd\u662f\u81ea\u7136\u6570\uff0c\u4e14\u5982\u679cx\u2208M\uff0c\u52198\uff0dx\u2208M\uff0c\u5219\u6ee1\u8db3\u8fd9\u6837\u6761\u4ef6\u7684\u96c6\u5408M\u7684\u4e2a\u6570\u4e3a( )(\u6ce8\uff1a\u81ea\u7136\u6570\u5305\u62ec0)
A.64 B.32 C.16 D.8\u6c42\u96c6\u5408{x\u2208Z| \u22642x\uff1c32}\u7684\u771f\u5b50\u96c6\u4e2a\u6570.
4. \u57281\uff5e120\u7684120\u4e2a\u81ea\u7136\u6570\u4e2d\uff0c\u7d20\u6570\u4e0e\u5408\u6570\u5404\u6709\u591a\u5c11\u4e2a\uff1f
5. \u5df2\u77e5M\uff1d{a\uff0ca\uff0bd\uff0ca\uff0b2d}\uff0cN\uff1d{a\uff0caq\uff0caq2}\uff0c\u4e14M\uff1dN\uff0c\u6c42q\u7684\u503c.
6. \u5728\u6570\u7406\u5316\u4e09\u79d1\u7ade\u8d5b\u8f85\u5bfc\u4e2d\uff0c\u9ad8\u4e0010\u300111\u300112\u73ed\u53c2\u52a0\u6570\u5b66\u8f85\u5bfc\u7684\u6709168\u4eba\uff0c\u53c2\u52a0\u7269\u7406\u8f85\u5bfc\u7684\u6709187\u4eba\uff0c\u53c2\u52a0\u5316\u5b66\u8f85\u5bfc\u7684\u6709155\u4eba\uff0c\u6570\u5b66\u3001\u7269\u7406\u4e24\u79d1\u90fd\u53c2\u52a0\u7684\u6709139\u4eba\uff0c\u6570\u5b66\u3001\u5316\u5b66\u4e24\u79d1\u90fd\u53c2\u52a0\u7684\u6709127\u4eba\uff0c\u7269\u7406\u3001\u5316\u5b66\u4e24\u79d1\u90fd\u53c2\u52a0\u7684\u6709135\u4eba\uff0c\u6570\u7406\u5316\u4e09\u79d1\u90fd\u53c2\u52a0\u7684\u6709102\u4eba\uff0c\u95ee\u8fd9\u4e09\u4e2a\u73ed\u603b\u5171\u6709\u591a\u5c11\u4eba\u81f3\u5c11\u53c2\u52a0\u4e86\u4e00\u79d1\u7684\u8f85\u5bfc\uff1f
\u89e3\uff1a\u6839\u636e\u5bb9\u65a5\u539f\u7406\uff0c\u81f3\u5c11\u53c2\u52a0\u4e00\u79d1\u8f85\u5bfc\u7684\u5b66\u751f\u4eba\u6570\u4e3a\uff1a
168\uff0b187\uff0b155\uff0d139\uff0d127\uff0d135\uff0b102\uff1d211
7. \u6c42\u8bc1\uff1a\u4efb\u610fn\uff0b1\u4e2a\u6574\u6570\u4e2d\uff0c\u603b\u6709\u4e24\u4e2a\u6574\u6570\u7684\u5dee\u80fd\u88abn\u6574\u9664\u3002
\u63d0\u793a\uff1a\u5229\u7528\u4f59\u6570\u6784\u9020n\u4e2a\u96c6\u5408\uff0c\u6839\u636e\u62bd\u5c49\u539f\u7406\uff0c\u81f3\u5c11\u6709\u4e24\u4e2a\u6574\u6570\u653e\u5728\u4e00\u4e2a\u96c6\u5408\u91cc\uff0c\u5b83\u4eec\u540c\u4f59\uff0c\u5b83\u4eec\u7684\u5dee\u4e00\u5b9a\u80fd\u88abn\u6574\u9664.
8. \u8bc1\u660e\uff1a\u82e5\u8d2d\u4e70\u8d85\u8fc717\u5343\u514b\uff08\u6574\u6570\u5343\u514b\uff09\u7684\u7cae\u98df\uff0c\u53ea\u75283\u5343\u514b\u548c10\u5343\u514b\u7684\u7cae\u7968\u652f\u4ed8\uff0c\u800c\u65e0\u9700\u8981\u627e\u8865\u3002
\u89e3\uff1a\u672c\u9898\u5176\u5b9e\u5c31\u662f\u8bc1\u660e\u5927\u4e8e17\u7684\u6574\u6570\u90fd\u80fd\u8868\u793a\u4e3a3m\uff0b10n\u7684\u5f62\u5f0f\uff0c\u5176\u4e2dm,n\u90fd\u662f\u975e\u8d1f\u6574\u6570.\u6ce8\u610f\u5230\uff1a\u5927\u4e8e17\u7684\u6574\u6570\u53ef\u4ee5\u5199\u62103k,3k\uff0b1,3k\uff0b2(k\u22656)\u7684\u5f62\u5f0f\uff0c\u800c3k\uff0b1\uff1d3(k\uff0d3)\uff0b10,3k\uff0b2\uff1d3(k\uff0d6)\uff0b10\u00d72\uff0c\u56e0\u6b64\u5b83\u4eec\u90fd\u80fd\u591f\u8868\u793a\u62103m\uff0b10n\u7684\u5f62\u5f0f\uff0c\u5176\u4e2dm,n\u90fd\u662f\u975e\u8d1f\u6574\u6570.
9. \u8bbeA\u662f\u6570\u96c6\uff0c\u6ee1\u8db3\u82e5a\u2208A\uff0c\u5219 \u2208A\uff0c\u4e141A.
\u2474\u82e52\u2208A\uff0c\u5219A\u4e2d\u81f3\u5c11\u8fd8\u6709\u51e0\u4e2a\u5143\u7d20\uff1f\u6c42\u51fa\u8fd9\u51e0\u4e2a\u5143\u7d20.
\u2475A\u80fd\u5426\u4e3a\u5355\u5143\u7d20\u96c6\u5408\uff1f\u5206\u522b\u5728\u5b9e\u6570\u96c6\u548c\u590d\u6570\u96c6\u4e2d\u8fdb\u884c\u8ba8\u8bba.
\u2476\u82e5a\u2208A\uff0c\u8bc1\u660e\uff1a1\uff0d \u2208A.
\u89e3\uff1a\u24742\u2208A  \uff0d1\u2208A  \u2208A  2\u2208A
\u2234 A\u4e2d\u81f3\u5c11\u8fd8\u6709\u4e24\u4e2a\u5143\u7d20\uff1a\uff0d1\u548c
\u2475\u5982\u679cA\u4e3a\u5355\u5143\u7d20\u96c6\u5408\uff0c\u5219a\uff1d
\u5373a2\uff0da\uff0b1\uff1d0
\u8be5\u65b9\u7a0b\u65e0\u5b9e\u6570\u89e3\uff0c\u6545\u5728\u5b9e\u6570\u8303\u56f4\u5185\uff0cA\u4e0d\u53ef\u80fd\u662f\u5355\u5143\u7d20\u96c6
\u4f46\u8be5\u65b9\u7a0b\u6709\u4e24\u4e2a\u865a\u6570\u89e3\uff1aa\uff1d i
\u6545\u5728\u590d\u6570\u8303\u56f4\u5185\uff0cA\u53ef\u4ee5\u662f\u5355\u5143\u7d20\u96c6\uff0cA\uff1d{ i}\u6216A\uff1d{ i}
\u2476a\u2208A  \u2208A  \u2208A\uff0c\u53731\uff0d \u2208A
10. \u8bbeS\u4e3a\u96c6\u5408{1\uff0c2\uff0c3\uff0c\u2026\u2026\uff0c50}\u7684\u4e00\u4e2a\u5b50\u96c6\uff0c\u4e14S\u4e2d\u4efb\u610f\u4e24\u4e2a\u5143\u7d20\u4e4b\u548c\u4e0d\u80fd\u88ab7\u6574\u9664\uff0c\u5219S\u4e2d\u5143\u7d20\u6700\u591a\u6709\u591a\u5c11\u4e2a\uff1f
\u5c06\u8fd950\u4e2a\u6570\u6309\u71677\u7684\u4f59\u6570\u5212\u5206\u62107\u4e2a\u96c6\u5408
A0={7,14,21,28,35,42,49}
A1={1,8,15,22,29,36,43,50}
A2={2,9,16,23,30,37,44}
A3={3,10,17,24,31,38,45}
A4={4,11,18,25,32,39,46}
A5={5,12,19,26,33,40,47}
A6={6,13,20,27,34,41,48}
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11. \u5df2\u77e5\u96c6\u5408A\u4e2d\u670910\u4e2a\u5143\u7d20\uff0c\u4e14\u6bcf\u4e2a\u5143\u7d20\u90fd\u662f\u4e24\u4f4d\u6574\u6570\uff0c\u8bc1\u660e\uff1a\u4e00\u5b9a\u5b58\u5728\u8fd9\u6837\u4e24\u4e2aA\u7684\u5b50\u96c6\uff0c\u5b83\u4eec\u4e2d\u6ca1\u6709\u76f8\u540c\u7684\u5143\u7d20\uff0c\u800c\u5b83\u4eec\u7684\u5143\u7d20\u4e4b\u548c\u76f8\u7b49.
\u89e3\uff1a\u8fd910\u4e2a\u5143\u7d20\u7684\u603b\u548cS\uff1c100\u00d710\uff1d1000
\u800cA\u7684\u5b50\u96c6\u603b\u5171\u6709210\uff1d1024\uff1e1000\uff1eS
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\u5982\u679cM\u3001N\u6ca1\u6709\u516c\u5171\u5143\u7d20\uff0c\u5219M\u3001N\u5c31\u662f\u6ee1\u8db3\u9898\u610f\u7684\u5b50\u96c6\uff0c\u547d\u9898\u5f97\u8bc1.
\u5982\u679cM\u3001N\u4e2d\u6709\u516c\u5171\u5143\u7d20\uff0c\u8bb0M\u2229N\uff1dQ,
\u8003\u67e5\u96c6\u5408M'\uff1dM\uff0dQ,N'\uff1dN\uff0dQ
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A.0 B.1 C.2 D.\uff0d1
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(2)a\u2208(1,+\u221e)
-2/(a-1)-2a=-2(a²-a+1)/(a-1)<0\u2192-2/(a-1)<2a
g(x)=f(x)+|x+2/(a-1)|,\u5206\u6bb5\uff1a
g(x)=-2x+2a-2/(a-1) x<-2/(a-1)
g(x)=2a+2/(a-1) -2/(a-1)\u2264x\u22642a
g(x)=2x-2a+2/(a-1) x>2a
\u2234g(x)\u7684\u6700\u5c0f\u503c=2a+2/(a-1)
g(x)2a+2/(a-1)\u22656

1.1. y=log3(x+1)/(x-1)+根号下(4-x^2)
因(x+1)/(x-1)>0 解得 x<-1或x>1
4-x^2≥0 x^2≤4 -2≤x≤2
故定义域为-2≤x<-1 或1<x≤2
当-2≤x<-1时
log3(x+1)/(x-1)∈(-∞, -1]
根号下(4-x^2)∈[0, √3)
所以y∈(-∞, -1]
当1<x≤2
log3(x+1)/(x-1)∈[1, +∞)
根号下(4-x^2)∈[0, √3)
所以y∈[1, +∞)
综上:值域y∈(-∞, -1]∪[1, +∞)
2.x=a
则a²+a-2013=0
a²=-a+2013

韦达定理a+b=-1

所以原式=-a+2013+2a+b
=a+b+2013
=-1+2013
=2012
3.1/8
sin^2(α/2)=(1-cosα)/2
cos^2(α/2)=(1+cosα)/2
推导:cos(2α)=cos(α+α)=cosαcosα-sinαsinα=cos^2(α)-sin^2(α)……①
在等式①两边加上1,整理得:cos(2α)+1=2cos^2(α)
将α/2代入α,整理得:cos^2(α/2)=(cosα+1)/2
在等式①两边减去1,整理得:cos(2α)-1=-2sin^2(α)
将α/2代入α,整理得:sin^2(α/2)=(1-cosα)/2
tan^2(α/2)=(1-cosα)/(1+cosα)
sin(α/2)=±[(1-cosα)/2]^(1/2)(正负由α/2所在象限决定)
cos(α/2)=±[(1+cosα)/2]^(1/2)(正负由α/2所在象限决定)
tan(α/2)=sinα/(1+cosα)=(1-cosα)/sinα=±[(1-cosα)/(1+cosα)]^(1/2)
推导:tan(α/2)
=sin(α/2) /cos(α/2)
=[2sin(α/2)cos(α/2] /2cos(α/2)^2
=sinα/(1+cosα)
=(1-cosα)/sinα
4.答案是:1-√3
5.延长AP交BC于D,延长CP交AB与E
AP=λAD=(2/5)AB+(1/5)AC
AD=xAB+yAC,x+y=1【共线定理,B,C,D共线】
得x/y=2,x+y=1,λ=x/(2/5),得λ=(3/5)
于是PD=(2/5)AD,S△PBC=(2/5)S△ABC
CP=μCE
CE=mCA+nCB,m+n=1【共线定理,E,A,B共线】
AP=(2/5)AB+(1/5)AC
AC+CP=(2/5)(AC+CB)+(1/5)AC
得CP=(2/5)CB+(2/5)CA
于是m=n=1/2,μ=(4/5)
于是CP=(4/5)CE,PE=(1/5)CE,S△PAB=(1/5)S△ABC
于是S△ABP/S△BCP=1/2
追问
不好意思哈,根据你的步骤我可以算出x=2/3,从而λ=5/3而不是λ=3/5,请问我错在哪里了?还有您写的x/y=2是根据题中条件向量AP=2/5向量AB+1/5向量AC,向量AB与向量AC前的系数比为2得出的吗?
回答
你算的是AD=λAP。
(2/5)ab+(1/5)AC=λxAB+λyAC
于是λx/λy=(2/5)/(1/5)=2
即x/y=2【由向量共线得出的】
追问
按您写的应该是AP=λAD吧,还有能不能告诉我 根据你的步骤我可以算出x=2/3,从而λ=5/3而不是λ=3/5,请问我错在哪里?
回答
xλ=2/5,x=2/3,得λ=3/5
提问者评价
按照你说的,真的成功了,好开心,谢谢你!
6.右边虚部为0,左边z+1的实部应为0
7.设(1+x+x^2)^150=(求和符号的先方是300,下头是k=0)∑Ckx^k,其中C0,∧,C300是常数,则(求和符号先方100,下头k=0)C3k=____
8.、随机选取正11边形的3个不同顶点,它们构成锐角三角形的概率为____盼高手详细解说剖析,谢谢!答案是:1/3
9.正三棱锥顶点P在底面ABC的投影D是等边三角形ABC的中心。
易知AD=根号3,从而 PD=根号(PA^2-AD^2)=根号3.
棱锥体积为 PD*三角形ABC面积/3=9/4.

延长AD交BC于E,则AE为BC的中垂线。连接PE,
易知AE=3*(根号3)/2,PE=(根号15)/2.
在PD上取一点Q,使该点到底面的距离和该点到棱面的距离相等,
即过Q点作面PBC的垂线QF,使QD=QF。
则QF必与PE相交于点F,且QF为内切圆的半径。
直角三角形PQF与PED相似,有 PQ/QF=PE/ED。 即PD/QF-1=PE/ED,
又ED=(根号3)/2, 得内切圆半径 QF=(根号3)/(1+根号5).
内切球的表面积 4πQF^2=3π(3-根号5)/2.
10.证明:f(x)对任意x,y属于R都有f(x+y)=f(x)+f(y)
令x=y=0,得f(0)=0
再令y=-x,得f(0)=f(x)+f(-x)
所以,f(-x)=-f(x),f(x)为奇函数
11.15.2k
12.an=3^n-1+a(n+1) 可得:a(n+1)-an=3^n-1
an-a(n-1)=3^(n-1)-1
a(n-1)-a(n-2)=3^(n-2)-1
......
a2-a1=3^1-1
以上式子累加:an-a1=3^(n-1)+3^(n-2)+...+3^1-(n-1)=3*(1-3^(n-1))/(1-3)-(n-1)=(3^n-3)/2-n+1
an=(3^n-3)/2-n+1+a1=(3^n-3)/2-n+2==(3^n+1)/2-n

OK了 qq1259545502

一楼不能再吊 祝获得Imo金牌
马一个 不过Lz你这个题量和0悬赏几乎不可能得到答案

都不很难,慢慢做

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