共查询到20条相似文献,搜索用时 140 毫秒
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第24届IMO第6题是:在△ABC中,a、b、c是三边长,求证:a2b(a-b)+b2c(b-c)+c2a(c-a)≥0.(1)文[1]指出了它的下述对偶形式:ab2(a-b)+bc2(b-c)+ca2(c-a)≤0,(2)并给出了统一的距离解释.即不等式(1)、(2)的几何解释为:三角形内Brocard点到内心的距离非负.受此启发,笔者研究了第6届IMO第2题:在△ABC中,a、b、c是三边长,求证: a2(b+c-a)+b2(a+c-b)+c2(a+b-c)≤3abc,(3)发现它也有如下的… 相似文献
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判定(一)(i)命题:若a,b,c∈R,且a≠0,b≠-1分式方程:cx-a+b=b-xx-a当b=a+c时,必有x=a为分式方程的增根。(i)例举:(1)1x-1+2=2-xx-1,(b=a+c即2=1+1),x=1是方程的增根。(2)3x+2+1... 相似文献
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与匹多不等式有关的一个等式方明(四川平昌二中635400)约定a,b,c,△和a′,b′,c′,△′分别表示△ABC和△A′B′C′的边长和面积,H=a′2(b2+c2-a2)+b′2(c2+a2-b2)+c′2(a2+b2-c2).著名的匹多不等式... 相似文献
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对于某些不等式证明题,我们若能根据其条件和结论,结合判别式的结构特征,通过构造二项平方和函数:f(x)=(a1x-b1)2+(a2x-b2)2+…+(anx-bn)2,由f(x)≥0,得Δ≤0,就可以使一些用一般方法处理较繁的问题,获得简捷、明快的证明.例1 已知a,b,c∈R+,求证:a2b+c+b2c+a+c2a+b≥a+b+c2.(第二届“友谊杯”国际数学邀请赛题)证 构造函数f(x)=(ab+cx-b+c)2+(bc+ax-c+a)2+(ca+bx-a+b)2=(a2b+c+b2c+a+… 相似文献
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一条件恒等式证明之我见徐鸿迟(江苏省泰州中学225300)考察这样的问题:已知a+b+c=abc,求证a(1-b2)(1-c2)+b(1-c2)(1-a2)+c(1-a2)(1-b2)=4abc.徐南昌在[1]中用数学审美的目光发现了下面的“证法”:... 相似文献
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贵刊文[1]通过构造恒等式 a2b+c+b2c+a+c2a+b-a+b+c2 =(a+b+c)(ab+c+bc+a+ca+b-32)巧妙地证明了著名不等式(1)、(2)的等价性:命题1 (1963年莫斯科竞赛题)设a、b、c∈R+,求证: ab+c+bc+a+ca+b≥32.(1)命题2 (第二届“友谊杯”国际数学竞赛题)设a、b、c∈R+,求证:a2b+c+b2c+a+c2a+b≥a+b+c2.(2)受其启发,我们可得更为一般的结论:设a、b、c∈R+,n∈N,则 anb+c+bnc+a+c… 相似文献
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从威森波克不等式的证明谈起武爱民(河南鹤壁四矿中学458010)威氏不等式:a2+b2+c243△(其中a,b,c和△分别为△ABC的边和面积).目前人们已发现了它的十多种证法,而且被加强为a2+b2+c243△+(a-b)2+(b-c)2+(c... 相似文献
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用asinx+bcosx=c有解的条件探求高考试题杜楚琼,陈建军(湖南邵东八中)高中《代数》上册第236页中指出:形如asinx+bcosx=c的三角方程(a、b不同时为零)有解的条件是我们不妨把△=a2+b2-c2称为上述三角方程的判别式.那么a2?.. 相似文献
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We consider certain pseudoconvex domains in C
n+1 and show that if the automorphism group is noncompact, then the domain is equivalent to
for some integerm ≥ 1.
Communicated by Steven Krantz 相似文献
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构造了一类步数为2(k+1)的次黎曼流形,给出其上连接原点和t轴上一点测地线的条数和相应测地线的长度,同时得到其中最短的测地线. 相似文献
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Andreas Brandt 《Mathematical Methods of Operations Research》1996,44(1):49-74
In this paper we investigate anN server loss system, where the input is a superposition of two types of traffics, namely of a renewal process and a Poisson process. The holding times of the two customer types are exponentially distributed with different parameters. For this model, denoted by
, we derive a numerical algorithm for computing the individual blocking (loss) probabilities. The analysis is given by constructing a two-dimensional embedded Markov chain and by using the intensity conservation principle as well as point process arguments. The results generalize those of Kuczura [8] and Willie [11]. Finally, for the
loss system we give a system of partial differential equations for the densities of the steady state distribution and discuss a special case. 相似文献
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关于指数丢番图方程a~x+b~y=c~z的Terai猜想 总被引:11,自引:2,他引:9
本文证明了:当a=|m(m4-10m2+)|,b=5m4-10m2+1,c=m2+1,其 中m是偶数时,如果m≥542,则方程ax+by=cz仅有正整数解(x,y,z)=(2,2,5). 相似文献
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Tomasz Beberok 《Journal of Geometric Analysis》2016,26(4):2883-2892
In this paper we obtain the closed forms of some hypergeometric functions. As an application, we obtain the explicit forms of the Bergman kernel functions for Reinhardt domains \(\{|z_3|^{\lambda } < |z_1|^{2p} + |z_2|^2, \ |z_1|^{2p} + |z_2|^2 < |z_1|^{p} \}\) and \(\{|z_4|^{\lambda } < (|z_1|^2 + |z_2|^2)^{p} + |z_3|^2, \ (|z_1|^2 + |z_2|^2)^{p} + |z_3|^2 < (|z_1|^2 + |z_2|^2 )^{p/2} \}\). 相似文献
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The cycle length distribution of a graph G of order n is a sequence (c1 (G),…, cn (G)), where ci (G) is the number of cycles of length i in G. In general, the graphs with cycle length distribution (c1(G) ,…,cn(G)) are not unique. A graph G is determined by its cycle length distribution if the graph with cycle length distribution (c1 (G),…, cn (G)) is unique. Let Kn,n+r be a complete bipartite graph and A lohtaib in E(Kn,n+r). In this paper, we obtain: Let s 〉 1 be an integer. (1) If r = 2s, n 〉 s(s - 1) + 2|A|, then Kn,n+r - A (A lohtain in E(Kn,n+r),|A| ≤ 3) is determined by its cycle length distribution; (2) If r = 2s + 1,n 〉 s^2 + 2|A|, Kn,n+r - A (A lohtain in E(Kn,n+r), |A| ≤3) is determined by its cycle length distribution. 相似文献
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本文研究一类带奇异项及临界指数的方程 利用纤维方法证明方程在满足一定条件下正解的存在性. 相似文献
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Let p(z) be a polynomial of degree at most n. In this paper we obtain some new results about the dependence of p(Rz)-βp(rz) + α (R+1/r+1)n-|β | p(rz) s on p(z) s for every α, β∈ C with |α|≤ 1, |β | ≤ 1, R > r 1, and s > 0. Our results not only generalize some well known inequalities, but also are variety of interesting results deduced from them by a fairly uniform procedure. 相似文献
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<正> 1.如果函數f(z)在包含實軸土某一區間的區域B中是正則的,f(z)在此實軸區間上取實值.在區域B的其餘地方f(z)與(z)同符號;即 相似文献