共查询到20条相似文献，搜索用时 62 毫秒
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徐邦腾 《黄冈师范学院学报》1990,(3)
设 R 是一个有单位元的交换环,maxspccR 是 R 的所有极大理想作成的集合.假设ζ是 R 的一个 Gabriel 拓扑,M∈maxspccR,则ζ_M={I_M|I∈ζ}是 R_M 的一个 Gabriel 拓扑.对任意 M∈maxspccR,设 C[M]是 R_M 的 Gabriel 拓扑ζ_M 的一个内射上生成子,则П_(M∈maxspccR)C[M]是内射 R—模。我们的问题是在什么条件下,П_(M∈maxspccR)C[M]是 R 的 Gabriel 拓扑ζ的一个内射上生成子.为叙述方便,我们称具有上述性质 相似文献
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性质1已知x,y,z∈R,且a,b,c∈R ,则axy byz czx≤α(x2 y2 z2)时,α的最小值是方程4x3-(a2 b2 c2)x-abc=0的正根.这是文[1]中,钱照平老师提出问题.证明:α(x2 y2 z2)-axy-byz-czx=α(x-ay2 αcz)2 4α24-αa2(y-42αbα2 -aac2z)2 [4α24-αc2-4(α2(b4αα 2-aca)22)]z2.设α> 相似文献
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第 31届西班牙数学奥林匹克第 2题是 :证明 :如果 ( x+ x2 + 1 ) ( y+ y2 + 1 )= 1 ,那么 x+ y=0 .文 [1 ]给出了此题的一种证法 ,本文再给出此题的两种换元证法 ,然后给出一个新命题 .证法 1 设 x=tanα,y=tanβ,其中 α,β∈ ( - π2 ,π2 ) ,则由条件知 ,( tanα+ secα) ( tanβ+ secβ) =1 ( sinα+ 1 ) ( sinβ+ 1 ) =cosαcosβ sinα+sinβ+ 1 =cos(α+β) 2 sinα+β2 cosα-β2 +1 =1 - 2 sin2 α+β2 sin α+β2 ( sin α+β2 +sinπ-α+β2 ) =0 sin α+β2 sin 2β+π4 ·cos2α-π4 =0 .又由 α,β∈ ( - π2 ,π2 ) ,知… 相似文献
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三角恒等式 :cosα cos(1 2 0°-α) cos(1 2 0° α) =0 ,sinα- sin(1 2 0°- α) sin(1 2 0° α) =0 .其中 α为任意角 .文 [1 ]、[2 ]先后给出了这两个恒等式的统一证法 .其实 ,笔者得以下证法更显朴素自然 ,简捷明快 !证明 记P=cosα cos(1 2 0°- α) cos(1 2 0° α) ,Q=sinα- sin(1 2 0°-α) sin(1 2 0° α) .则 P2 Q2 =3 2 [cosαcos(1 2 0°-α)- sinαsin(1 2 0°- α) ] 2 [cosαcos(1 2 0° α) sinαsin(1 2 0° α) ] 2 [cos(1 2 0°- α)·cos(1 2 0° α) - sin(1 2 0°- α) sin(1 2 0° … 相似文献
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题目 已知 3ｓｉｎ2 α +2ｓｉｎ2 β =2ｓｉｎα,求ｓｉｎ2 α +ｓｉｎ2 β的取值范围 .错解 ∵ 3ｓｉｎ2 α+2ｓｉｎ2 β=2ｓｉｎα,∴ｓｉｎ2 α+ｓｉｎ2 β =ｓｉｎ2 α +12 ( 2ｓｉｎα -3ｓｉｎ2 α) =-12 ｓｉｎ2 α+ｓｉｎα =-12 (ｓｉｎα-1 ) 2 +12 .∵ｓｉｎα∈ [-1 ,1 ],∴ｓｉｎ2 α +ｓｉｎ2 β∈ -32 ,12 .剖析 在上述求解过程中 ,已注意到ｓｉｎα取值范围 :-1 ≤ｓｉｎα≤ 1 ,但是还没有注意到题设条件对ｓｉｎα的取值限制 .事实上 ,由 3ｓｉｎ2 α+2ｓｉｎ2 β=2ｓｉｎα ,得ｓｉｎ2 β=12 ( 2ｓｉｎα-3ｓ… 相似文献
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主要证明了如下结果:(1) 如果X=∏α∈ΛXα是|Λ|-仿紧空间,则X是meso紧的当且仅当(A)F∈[Λ]＜ω,∏α∈FXα是meso紧的;(2)如果X=∏i∈ωXi是可数仿紧的,则下列三条件等价:X是meso紧的;(A)F∈[Λ]＜ω,∏α∈FXα是meso紧的;(A)n∈ω,∏isnXi是meso紧的. 相似文献
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杨先义 《中学数学教学参考》2008,(1):121-121
文[1]定义了Z数：对P∈N,P^2可从某处截断,分为M1、M2,如／M1-M2/=P,则称P为Z数.文[1]阐明了对n∈N,10^n、10^n＋1必为Z数（可谓平凡Z数）.此外,还找到非平凡Z数：3位数的有2个：287,364;4位数的有4个：1078,1096,1287,1364. 相似文献
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文[1]、[2]、[3]、[4]研究了在约束条件Ax~2+Dxy+Cy~2=M下,求函数ω=Ax~2+Bxy+Cy~2(A,B,C,D,M∈R)的最值、值域.本文给出该问题的另一种解法,即二元均值不等式的变式-(a~2+b~2)/2≤ab≤(a~2+b~2)/2(a,b∈R) 相似文献
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理科第 ( 1 7)题 :已知 sin2 2 α sin2 αcosα- cos 2α=1 ,α∈ ( 0 ,π2 ) ,求 sinα,tanα的值 .解法 1 sin2 2α sin 2αcosα- cos 2α=1 sin2 2α sin 2αcosα- 2 cos2 α=0 ( sin 2α 2 cosα) ( sin2 α- cosα) =0 sinα=12 ,α= π6 ,tanα=33.解法 2 sin2 2 α sin2 αcosα- cos2 α=1 2 sin2 α sinα- 1 =0 ( sinα 1 ) ( 2 sinα- 1 ) =0 sinα=12 ,α=π6 ,tanα=33.图 1理科第 ( 1 8)题 :如图 1 ,正方形ABCD,ABEF的边长都是 1 ,而且平面 ABCD,ABEF 互相垂直 ,点 M在 AC上移动 ,点 N在 BF上移… 相似文献
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文[1]谈了椭圆焦点三角形内心和旁心的轨迹方程,本文进一步谈双曲线焦点三角形内心和旁心的轨迹方程.设M点是双曲线xa22-yb22=1(a>0,b>0)上一点,F1(-c,0),F2(c,0)是双曲线的两个焦点,称三角形M F1F2为双曲线的焦点三角形.引理(1)设∠M F1F2=α,∠M F2F1=β,M点在双曲线右支上,则t a n2α?c o t2β=c-ac a;M点在双曲线左支上,则t a n2α?c o t2β=cc -aa.引理(2)(如图1)设M(x0,y0),△M F1F2的内心为K,连M K并延长M K交x轴于N点,则N点的横坐标xN=xa02.证明:(1)当M点在双曲线右支上时,鸐F1?s i nβ=鸐s i nFα2?s i n?(Fα1F 2?β… 相似文献
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《海外英语》2007,(5):10-11
Many college freshmen arrive woefully unprepared to do college work, and as disadvantaged populations continue to grow, the share of the American work force that has made it through college is expected to plummet. Many experts blame that educational failure not just on high schools but also on colleges. School & College, a special report by The Chronicle, looks at efforts to fix the system. What reforms would better prepare students for college? What should schools and colleges be doing differently? How should state and federal officials help? 相似文献
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《海外英语》2007,(4):36
There are numbers of crossroads on our long and unpredictable life journey where we totally have no idea about which direction to choose. No matter what our decision is, we should not turn back, but face the music and go ahead instead. I am this kind of girl who always does try without regretting, one example is how I dealt with my love. 相似文献
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Migration occurs behind a variety of reasons and has a great effect on the whole world. People may migrate in order to improve their economic situation, or in order to escape civil strife, persecution, and environmental disasters. The impact of migration is complex, bringing both benefits anddisadvantages. This paper briefly talks about the causes of migration, the allocation of benefits, and the ways in which individual countries and the international community deal with this important subject. 相似文献
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Sister Carrie is one of the most controversial characters in American literature.Thought as a "fallen woman" firstly,she was defined as a "new woman" by some critics later. However, by digging into the motivaton behind the whole process of Carrie's "success", the relationship between Carrie and her creator (the author), the social conditions of then American, it can be found that Carrie has never been free-standing on her thought and she has never found her real-sdf even after becoming a famous actress. In a society dominated by mass consumerism Carrie is only an adherent of her own desires. She also is a representative of all those country girls flooded into cities, a symbol and a sacrifice of the urbanization of America in a time countryside was overcome by cities. 相似文献
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RUZBEHANI Mohsen 《重庆大学学报(英文版)》2004,3(1):20-25
1.IntroductionOne-cyclecontrolmethod,whichwasproposedaboutonedecadeago[1],hasbecomeanattractivemethodinspecialfieldssuchaspowerfactorcorrection[2-6],switchingamplifiers[7,8],etc.Themainideaofthiscontrollerisbasedonintegrationofdiodevoltageinone-cycleandforcesittobeexactlyequaltothereferencevalue.Themainadvantageofthiscontrollerisitsrealtimeabilitytorejectthevariationofinputvoltage[1].Despitethisgreatability,ithasnogoodperformancesinrejectingofloaddisturbanceandfollowingreferencecommands.Espec… 相似文献
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唐芳 《读与写:教育教学刊》2008,5(6)
During interview with English majors, it was found that a majority of students expressed disappointment with their experience of English literature classes. Students' dislike of learning English literature appears to be the major problem of teaching English literature in China. Reader response approach is a good way to solve this problem. In this paper, I explain the rationale behind the reader response approach, such as its origin, definition, features, assumptions, and strategies. Then I illustrate how reader response approach works as a teaching strategy by presenting several teaching models. At last I evaluate its usefuiness for teachers and present the advantages of reader response approach in Chinese context. 相似文献