2010考研数学(一)真题及答案Created with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please visit: :products.asposewordsPAGE Created with an evaluation copy of As
You have to believe there is a way. The ancients said: the kingdom of heaven is trying to enter. Only when the reluctant step by step to go to it s time must be managed to get one step down only h
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Evaluation Only. Created with Aspose.Words. Copyright 2003-2022 Aspose Pty Ltd. 2010年考研数学一真题及答案Created with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please v
2010考研数学(一)真题及参考答案一选择题(1)极限( C)A1 B C D【详解】 (2)设函数由方程确定其中F为可微函数且则( B)A B C D【详解】 等式两边求全微分得: 所以有 其中代入即可(3)设是正整数则反常积分的收敛性( D )(A)仅与的取值有关 (B)仅与有关(C)与都有关 (D)都无关【详解】:显然
Created with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please visit: :products.asposewords
Created with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please visit: :products.asposewords2000年全国硕士研究生入学统一考试数学(一)试卷一填空题(本题共5小题每小题3分满分15分.把答案填在题中横线上)(1)=
Created with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please visit: :products.asposewordsPAGE Created with an evaluation copy of Aspose.Words. To d
Created with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please visit: :products.asposewordsCreated with an evaluation copy of Aspose.Words. To discover t
2002年考研数学一试题答案与解析一填空题(1)【分析】原式(2)【分析】方程两边对两次求导得①②以代入原方程得以代入①得再以代入②得(3)【分析】这是二阶的可降阶微分方程.令(以为自变量)则代入方程得即(或但其不满足初始条件).分离变量得积分得即(对应)由时得于是积分得.又由得所求特解为(4)【分析】因为二次型经正交变换化为标准型时标准形中平方项的系数就是二次型矩阵的特征值所以是的特征值.
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