565 lines
324 KiB
SQL
565 lines
324 KiB
SQL
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INSERT INTO `网页数据` VALUES ('1','1','1','全等三角形是初中数学一个重难点内容，相信很多同学和家长对下面这几条判定并不陌生，它们被大家俗称为“4+1”，4条是适用于所有三角形，1条是适用直角三角形。'),('1','2','2','1.jpg'),('1','3','1','很多同学，或者家长小时候在学这部分知识时，一定都背过，但是当问到“为什么这4条能判定三角形全等”时，却很少能回答出来。如果只靠死记硬背这些判定，就只能停留在“是什么”上。'),('1','6','1','对于“为什么”该怎么学？掌握这两个字——梳理'),('1','5','1','我们来一一梳理这些判定规则，帮助大家理解和记忆。'),('1','7','1','①“边边边”最好理解，因为三角形具有稳定性。也就是三角形的三条边长度确定以后，这个三角形就固定了。'),('1','4','1','②“边角边”的本质是一种动态变化。比如我们把两条边想象成是时针和分针在走，它们在开开合合；当喊停的时候，时针和分针不动了，其实也就是它们组成的角也固定住了。两条边长度不变，角也固定了，接下来只需要把第三条边连上。'),('1','8','2','2.jpg'),('1','9','1','③ 接下来是“角角边”和“角边角”。我们先来思考，三角形的角度决定了三角形的什么？'),('1','10','1','比如我们画两个三角形，它们的角度都是100°、50°、30°，我们会发现它们的形状是一样的。'),('1','11','2','3.jpg'),('1','12','1','所以角度决定了三角形的形状，而且只要有两个角相等了，它们的形状就是一样的。因为三角形三个角加起来是180°，有两个角相等了，另外一个也是相等的。'),('1','13','1','接下来再来思考，当两个三角形形状一样，边决定了三角形的什么？边的长度决定了三角形的大小。'),('1','14','1','\r\n所以当两个三角形，两个角相等决定了它们形状一样，再来任意一条边相等，这两个三角形就是全等三角形。'),('1','15','1','知道“为什么”之后，再来看“怎么做”，还是用一道题目来举例。'),('1','16','2','4.jpg'),('1','17','1','根据判定规则，我们记住这六个字，就可以解决这道题：共顶点，红配蓝（也可以其他两个颜色）'),('1','18','1','把公共顶点A的这两个三角形里面，相等的线段用红蓝两种颜色表示'),('1','19','2','5.jpg'),('1','20','1','画完之后，就变得很明显了，我们再把第三条边用相应的颜色连上，答案就出来了。'),('1','21','2','6.jpg'),('1','22','1','从“是什么”、“为什么”到“怎么做”是初中几何学习的核心方法，在学习几何的各个板块时，都要让孩子问自己，这三点是否真正学明白了。'),('2','1','1','在初中数学的考试当中，解所有题型都需要用到公式，公式的重要性不言而喻！为了更好的帮助孩子们记住初中数学常用的公式、定理，在此老师给大家总结了初中三年所有的公式及定理！这份资料十分珍贵，每次考试都能用的上！'),('2','2','2','1.jpg'),('2','3','2','2.jpg'),('2','4','2','3.jpg'),('2','5','2','4.jpg'),('2','6','2','5.jpg'),('2','7','2','6.jpg'),('2','8','2','7.jpg'),('2','9','2','8.jpg'),('2','10','2','9.jpg'),('2','11','2','10.jpg'),('2','12','2','11.jpg'),('2','13','2','12.jpg'),('2','14','2','13.jpg'),('2','15','2','14.jpg'),('2','16','2','15.jpg'),('2','17','2','16.jpg'),('2','18','2','17.jpg'),('2','19','2','18.jpg'),('2','20','2','19.jpg'),('2','21','2','20.jpg'),('2','22','2','21.jpg'),('3','1','1','家长们先来个<span style=\"color: red;\">#不生气挑战#</span>，点击观看下方视频，看看你能忍到第几个？'),('3','2','3','2.mp4'),('3','3','1','怎么样？有感触吧！平时孩子作业时，咱们辅导孩子是不是也和视频里一样鸡飞狗跳？情绪经常崩溃，做题全靠生活经验，辅导孩子他还听不懂......'),('3','4','1','很多家长百思不得其解：明明该教的也都教了，孩子也很努力，为什么孩子在数学上就是不“<u>开窍</u>”呢？'),('3','5','1','这里就涉及到我们常说的一个概念——“数学思维”。这个概念听起来高大上，理解起来却很模糊。但它往往是影响孩子数学成绩的重要因素，今天就给大家解读一下这个玄乎的“数学思维”！'),('3','6','1','数学思维是何方神圣？'),('3','7','1','简单来说，「能够用数学的观点去思考问题和解决问题的能力」就是数学思维。'),('3','8','1','孩子数学跟不上，多数情况是因为数学思维不好！打个比方，假如每道数学题都是一片沙漠，孩子是迷路的旅人需要找到正确的方向，那么数学思维就是带领旅人走出沙漠的导航。'),('3','9','1','如果没有数学思维，孩子就没有一条明确的解题路线，四处试探的结果只能能是解题失败，还非常容易陷入“死记硬背、做题不会”的困境。'),('3','10','2','3.jpg'),('3','11','1','数学思维如何培养？'),('3','12','1','1、让孩子试着“讲题”'),('3','13','1','会做不一定会讲，孩子做完题后，家长不妨鼓励孩子开口讲解一下某道题，当一当家长的小老师。'),('3','14','1','孩子要想讲给家长听，就要思考、分析、整理这道题的解题思路，然后有条理地讲出来，这个知识就牢牢地成为他自己的知识，并且能够举一反三、触类旁通。'),('3','15','1','家长激励孩子多讲题，也为他建立了学习的信心和成就感。孩子能开口讲题、讲思路，这就是最好的数学思维训练模式。'),('3','16','2','4.jpg'),('3','17','1','2、抽象算式“具象化”'),('3','18','1','把抽象的数学实践到具体生活中，可以帮助孩子更容易理解。比如去超市买东西，通过标签价格、数量让孩子帮忙算算有没有超出预算等。'),('3','19','1','解题也一样，比如锯木头问题，锯四段需要6分钟，6不是4的倍数怎么算呢？这时候可以让孩子把这根木头画出来，分成4段就会发现其实只锯了3次，那问题就迎刃而解了。'),('3','20','1','转化思路，通过具象思维帮助孩子理解，善于发现这些隐藏在生活中的数学思维，学数学就会变成一件轻松有趣的事。'),('3','21','2','5.jpg'),('3','22','1','因此，学好数学需要抓住思维培养的关键期3-12岁，这个时期孩子的大脑快速发育，思维活跃，可塑性强，最有利于数学思维的培养！而数学常用的16种思维方法：分类思想、转化思想、假设思想、类比思想、极限思想、可逆思想等。'),('3','23','1','想要在小学这个关键时期，培养好这些数学思维方法，最直接有效的方法，就是选择一位资深老师，用系统的教材，通过科学的系统学习，才能构建完整的知识体系。'),('4','1','1','追本溯源化繁为简'),('4','2','1','目有千万而纲为一，枝叶繁多而本为一。纲举则目张，执本而末从。如果只在细枝末节上下功夫，费了力气却讨不了好。学习就是不断地归一，最终以一心一理贯通万事万物，则达自由无碍之化境矣（呵呵，这境界有点高，慢慢来）。'),('4','3','1','关于几何最值问题研究的老师很多，本人以前也有文章论述，本文在此基础上再次进行归纳总结，把各种知识、方法、思想、策略进行融合提炼、追本溯源、认祖归宗，以使解决此类问题时更加简单明晰。'),('4','5','1','所有问题的老祖宗只有两个：①[定点到定点]：两点之间，线段最短；②[定点到定线]：点线之间，垂线段最短。'),('4','5','2','1.gif'),('4','7','2','2.gif'),('4','8','1','由此派生：<br>③[定点到定点]：三角形两边之和大于第三边；<br>④[定线到定线]：平行线之间，垂线段最短；<br>⑤[定点到定圆]：点圆之间，点心线截距最短（长）；<br>⑥[定线到定圆]：线圆之间，心垂线截距最短；<br>⑦[定圆到定圆]：圆圆之间，连心线截距最短（长）。'),('4','9','1','余不赘述，下面仅举一例证明：'),('4','10','1','[定点到定圆]：点圆之间，点心线截距最短（长）。'),('4','11','1','已知⊙O半径为r，AO=d，P是⊙O上一点，求AP的最大值和最小值。'),('4','12','2','3.png'),('4','13','1','证明：'),('4','14','1','由“两点之间，线段最短”得AP≤AO+PO，AO≤AP+PO，'),('4','15','1','得d-r≤AP≤d+r，AP最小时点P在B处，最大时点P在C处。'),('4','16','1','即过圆心和定点的直线截得的线段AB、AC分别最小、最大值。'),('4','17','1','(可用“三角形两边之和大于第三边”，其实质也是由“两点之间，线段最短”推得）。'),('4','18','1','上面几种是解决相关问题的基本图形，所有的几何最值问题都是转化成上述基本图形解决的。'),('4','19','1','二、考试中出现的问题都是在基本图形的基础上进行变式，如圆与线这些图形不是直接给出，而是以符合一定条件的动点的形式确定的；再如过定点的直线与动点所在路径不相交而需要进行变换的。'),('4','20','1','类型分三种情况：'),('4','21','1','（1）直接包含基本图形；'),('4','22','1','（2）动点路径待确定；'),('4','23','1','（3）动线（定点）位置需变换。'),('4','24','1','（一）直接包含基本图形。'),('4','25','1','例1：'),('4','26','1','在⊙O中，圆的半径为6，∠B=30°，AC是⊙O的切线，则CD的最小值是 。'),('4','27','2','4.gif'),('4','28','1','简析：由∠B=30°知弧AD一定，所以D是定点，C是直线AC上的动点，即为求定点D到定线AC的最短路径，求得当CD⊥AC时最短为3。'),('4','29','1','（二）动点路径待确定。'),('4','30','1','例2：'),('4','31','1','如图，在△ABC中，∠ACB=90°，AB=5，BC=3，P是AB边上的动点（不与点B重合），将△BCP沿CP所在的直线翻折，得到△B′CP，连接B′A，则B′A长度的最小值是 。'),('4','32','2','5.gif'),('4','33','1','简析：A是定点，B\'是动点，但题中未明确告知B\'点的运动路径，所以需先确定B\'点运动路径是什么图形，一般有直线与圆两类。此题中B\'的路径是以C为圆心，BC为半径的圆弧，从而转化为定点到定圆的最短路径为AC-B\'C=1。'),('4','34','1','例3：'),('4','35','1','在△ABC中，AB=AC=5，cos∠ABC=3/5，将△ABC绕点C顺时针旋转，得到△A\'B\'C，点E是BC上的中点，点F为线段AB上的动点，在△A\'B\'C绕点C顺时针旋转过程中，点F的对应点是F\'，求线段EF\'长度的最大值与最小值的差。'),('4','36','2','6.png'),('4','37','1','简析：E是定点，F\'是动点，要确定F\'点的运动路径。先确定线段A\'B\'的运动轨迹是圆环，外圆半径为BC，内圆半径为AB边上的高，F\'是A\'B\'上任意一点，因此F\'的运动轨迹是圆环内的任意一点，由此转化为点E到圆环的最短和最长路径。'),('4','38','2','7.gif'),('4','39','2','8.png'),('4','40','2','9.png'),('4','41','1','E到圆环的最短距离为EF2=CF2-CE=4.8-3=1.8，'),('4','42','1','E到圆环的最长距离为EF1=EC+CF1=3+6=9，其差为7.2。'),('4','43','1','（三）动线（定点）位置需变换。'),('4','44','1','线段变换的方法：'),('4','45','1','（1）等值变换：翻折、平移；'),('4','46','1','（2）比例变换：三角、相似。'),('4','47','1','【翻折变换类】典型问题：“将军饮马”。'),('4','48','1','例4：'),('4','49','1','如图，∠AOB=30°，点M、N分别是射线OA、OB上的动点，OP平分∠AOB，且OP=6，当△PMN的周长最小值为 。'),('4','50','2','10.gif'),('4','51','1','简析：动线段（或定点）应居于动点轨迹的两侧，本题的三条动线段PM、MN、PN在OA、OB的内侧。所以本题的关键是把定线段变换到动点轨迹的两侧，从而把三条动线段PM、MN、PN转化为连接两点之间的路径。如图，把点P分别沿OA、OB翻折得P1、P2，△PMN的周长转化为P1M+MN+P2N，这三条线段的和正是连接两个定点P1、P2之间的路径，从而转化为求P1、P2两点之间最短路径，得△PMN的周长最小值为线段P1P2＝OP＝6。'),('4','52','1','例5：'),('4','53','1','如图，在锐角△ABC中，AB=4，∠BAC=45°，∠BAC的平分线交BC于点D，M、N分别是AD和AB上的动点，则BM+MN的最小值是 。'),('4','54','2','11.gif'),('4','55','1','简析：本题的问题也在于动线段BM、MN居于动点轨迹AD的同侧，同样把点N沿AD翻折至AC上，BM+MN＝BM+MN\'，转化为求点B到直线AC的最短路径，即BN\'⊥AC时，最小值为2√2。'),('4','56','1','【平移变换类】典型问题：“造桥选址”。'),('4','57','1','例6：'),('4','58','1','如图，m、n是小河两岸，河宽20米，A、B是河旁两个村庄，要在河上造一座桥，要使A、B之间的路径最短应该如何选址（桥须与河岸垂直）？'),('4','59','2','12.png'),('4','60','1','简析：桥长为定值，可以想像把河岸m向下平移与n重合，同时把点A向下平移河宽，此时转化成n上的一点到A、B的路径之和最短，即转化为定点A\'到定点B的最短路径。如下图：'),('4','61','2','13.gif'),('4','62','1','思路是把动线AM平移至A\'M，A\'N+BN即转化为求定点A\'与定点B之间的最路径。本题的关键是定长线段MN把动线段分隔，此时须通过平移把动线段A\'N、BN变为连续路径，也可以把点B向上平移20米与点A连接。'),('4','63','1','例7：'),('4','64','1','如图，CD是直线y=x上的一条定长的动线段，且CD=2，点A（4，0），连接AC、AD，设C点横坐标为m，求m为何值时，△ACD的周长最小，并求出这个最小值。'),('4','65','1','解析：两条动线段AC、AD居于动点所在直线的两侧，不符合基本图形中定形（点线圆）应在动点轨迹的两侧。首先把AC沿直线CD翻折至另一侧，如下图：'),('4','66','2','14.png'),('4','67','1','现在把周长转化为A\'C+CD+AD，还需解决一个问题：动线段A\'C与AD之间被定长线段CD阻断，动线段必须转化成连续的路径。同上题的道理，把A\'C沿CD方向平移CD的长度即可，如下图。'),('4','68','2','15.png'),('4','69','1','现在已经转化为A\'\'D+AD的最短路径问题，属定点到定点，当A\'\'D与AD共线时A\'\'D+AD最短，即为线段AA\'\'的长。'),('4','70','2','16.png'),('4','71','1','【三角变换类】典型问题：“胡不归”。'),('4','72','1','例8：'),('4','73','1','如图，A地在公路BC旁的沙漠里，A到BC的距离AH＝2√3，AB＝2√19，在公路BC上行进的速度是在沙漠里行驶速度的2倍。某人在B地工作，A地家中父亲病危，他急着沿直线BA赶路，谁知最终没能见到父亲最后一面，其父离世之时思念儿子，连连问：“胡不归，胡不归……！”（怎么还不回来），这真是一个悲伤的故事，也是因为不懂数学而导致的。那么，从B至A怎样行进才能最快到达？'),('4','74','2','17.png'),('4','75','1','简析：BP段行驶速度是AP段的2倍，要求时间最短即求BP/2+AP最小，从而考虑BP/2如何转化，可以构造含30°角利用三角函数关系把BP/2转化为另一条线段。如下图，作∠CBD=30°，PQ⊥BD，得PQ=1/2BP，由“垂线段最短”知当A、P、Q共线时AP+PQ＝AQ\'最小。'),('4','76','2','18.png'),('4','77','1','【相似变换类】典型问题：“阿氏圆”。'),('4','78','1','“阿氏圆”：知平面上两点A、B，则所有满足PA/PB=k且不等于1的点P的轨迹是一个圆，这个轨迹最先由古希腊数学家阿波罗尼斯发现，故称阿氏圆，如下图所示，其中PO：BO＝AO：PO＝PA：PB＝k。'),('4','79','2','19.gif'),('4','80','1','例9：已知A(-4，-4)、B(0, 4)、C(0, -6)、 D(0, -1)，AB与x轴交于点E，以点E为圆心，ED长为半径作圆，点M为⊙E上一动点，求 1/2AM+CM 的最小值。'),('4','81','2','20.png'),('4','82','1','简析：本题的主要问题在于如何转化1/2AM，注意到由条件知，在M的运动过程中，EM：AE＝1：2保持不变，从而想到构造相似三角形，使之与△AEM的相似比为1：2，这样便可实现1/2AM的转化，如下图取EN：EM＝1：2，即可得△EMN∽△EAM，再得MN=1/2AM，显然，MN+CM的最小值就是定点N、C之间的最短路径。'),('4','83','2','21.png'),('4','84','1','之后便是常规方法先求N点坐标，再求CN的长。'),('4','85','2','22.png'),('4','86','1','【解法大一统】'),('4','87','1','万法归宗：路径成最短，折线到直线。'),('4','88','1','（所求路径在一般情况下是若干折线的组合，这些折线在同一直线上时即为最短路径）'),('4','89','1','基本图形：动点有轨迹，动线居两边。'),('4','90','1','（动点轨迹可以是线或圆，动线指动点与定点或定线、定圆的连线，动线与折线同指）'),('4','91','1','核心方法：同侧变异侧，分散化连续。'),('4','92','1','（动线在同侧，要变为异侧，一般用翻折、三角、相似的方法构造；动折线被定长线段分散时需化为连续折线，一般用平移的方法构造，如造桥选址问题）'),('4','93','1','下图是构造完成的目标图形：'),('4','94','2','23.png'),('4','95','1','再举2例说明上述规律的运用方法：'),('4','96','1','1.'),('4','97','1','如图，菱形ABCD中，∠A=60°，AB＝3，⊙A、⊙B的半径为2和1，P、E、F分别是CD、⊙A、⊙B上的动点，则PE+PF的最小值为 。'),('4','98','2','24.png'),('4','99','1','思考方法如下图所示：'),('4','100','2','25.png'),('4','101','1','2.'),('4','102','1','菱形ABCD中，∠BAC=60°，P是AC上的动点，求BP+1/2AP的最小值。'),('4','103','2','26.png'),('4','104','1','思考方法如下图所示：'),('4','105','2','27.png'),('4','4','1','一、基本图形'),('5','1','2','1.png'),('6','1','2','1.png');
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