这道题还是用均值不等式来解好

题　已知a、b、c,x、y、z是实数 ,求a2 +b2 +c2 =1,x2 + y2 +z2 =9,求ax+by +cz的最大值 .该题是常出现在一些课外资料及杂志上的题目 ,学生在解题时往往用均值不等式来解 ,但由于忽视了a2 +b2 +c2 ≠x2 + y2 +z2 而导致解题错误 .为此 ,一些杂志上采用柯西不等式来进行求解 ,但学生对柯西不等式知之甚少 ,若用这种方法 ,学生难以掌握和理解 ,而且也不符合大部分学生的实际情况 .笔者认为 ,在解题中只要对该题的已知式进行适当的变形 ,仍可用均值不等式来解 ,现分析解答如下 :分析　该题的问题是由于a2 +b2 +c2 ≠x2 +y2 +z2 而导致不能用均…     

一类用均值不等式求最大值问题的推广

有时我们会遇到这样一类求极值的问题,如：已知x、y、zeR“且Zx＋3y＋sz一民求XyZ的最大值。根据均值不等式我们知道,当函数式中各项之和为定位时,其积有最大值。因此可用均值不等式求之。解：”．”2X＞0,3y＞0,SZ＞0。且ZX＋3y＋SZ一6由均值不等式得这本身是一个小问题,如果我们把这个小题大做,就可做出如下的文章。将问题做如下推广：推广1：已知x、y、z、a、b、c6R”且ax＋by＋cz—K（K为定值）,求xyz的最大值。解：同上解法得推广2：已知x、v、…、z、a、b…、c6R”且ax＋by＋…＋cz—K（K为定值）。求x·y…Z的最大值…     

向量与其他数学知识的交汇

向量具有数字化的形式,同时又具有形象化的特征,故成为联系多项数学知识的媒介.一、与代数的交汇【例1】设实数x、y、z、a、b、c满足条件:(x2 y2 z2)(a2 b2 c2)=(ax by cz)2,求证ax=by=cz.证明:设m=(x,y,z),n=(a,b,c),且m与n的夹角为θ.∵m·n=｜m｜·｜n｜cosθ,m·n=ax by cz∴ax by cz=x2 y2 z2·a2 b2 c2cosθ由已知得cosθ=±1,即θ=0或π.∴m∥n由向量平行充要条件是ax=by=cz.评注:在等式证明中,利用数量积公式,建立数形对应关系,从而问题得解.【例2】已知a,b,c,d∈R,求函数f(x)=(x a)2 b2 (x-c)2 d2的最小值.解:设m=(x a,b),n=(c-x,…     

一道最值问题的推广

文[1]利用均值不等式给出一道最值问题的通解(法一),并将该问题作了进一步的推广;文[2]用向量法对该问题及其推广进行解答(法二).本文将应用空间几何知识和柯西不等式,给出该问题及其推广的另外两种解法(法三,法四). 文[1]的问题及其推广是: 问题 已知a,b,c,x, y,z 是实数,a2 b2 c2=1, x2 y2 z2 = 9 ,求ax by cz 的最大值. 问题推广 已知ai,bi(i =1,2,L,n)且∑an n n 2 = p, 2 i ∑b i = q ,求 aibi 的最大值. ∑ i=1 i=1 i=1 …     

巧用待定系数法求多元最值问题

<正>待定系数法是数学的一个重要方法,其应用比较广泛,如多项式的因式分解,求函数的解析式和曲线的方程,求微分方程的级数形式的解等.本文例谈利用待定系数法求多元最值,希望能够抛砖引玉.先从一道题的错解谈起.引例已知实数a、b、x、y满足:a²+b2+b²=1,x2=1,x²+y2+y²=9,则ax+by的最大值为_.错解因为ax≤(a2=9,则ax+by的最大值为_.错解因为ax≤(a²+x2+x²)/2,by≤(b2)/2,by≤(b²+y2+y²)/2,当且仅当a=x,b=y时取等号.     

巧解最值问题

<正>二维形式的柯西不等式:若a,b,c,d都是实数,则(a2+b2)(c2+d2)≥(ac+bd)2,当且仅当ad=bc时,等号成立.上述不等式可以变形为:|ac+bd|a2+b%2姨≤c2+d%2姨,不等式的左边可以看成点(c,d)到直线ax+by=0的距离,当不等式的右边为定值时,左边有最大值.利用柯西不等式及其变形可以巧妙地解决如下最值问题.例1:求椭圆C:x216+y212=1上的点到直线l:x-2y=0的距离     

一道国家队培训题的另解

题目已知实数a、b、c、x、y、z满足（a＋b＋c）（x＋y＋z）=3，（a^2＋b^2＋c^2）（x^2＋y^2＋z^2）=4．求证：ax＋by＋cz≥0．     

一类基本不等式题型的最值

本文通过具体例题总结了基本不等式求一类题型（x+y）（a/x+b/y）（x,y,a,b都是正数）的最值.苏教版必修五给出了基本不等式的形式:ab^1/2≤（a+b）/2（a≥0,b≥0）,当且仅当a=b时取等号,其变形形式有a+b≥2ab^1/2基本不等式的一个运用就是求最值:①当a≥0,b≥0时,若和a+b为定值P,则积ab有最大值ab≤p²/4,当且仅当a=b时取等号;②当a≥0,b≥0时,若积ab为定值S,则和a+b有最小值a+b≥2S^1/2,当且仅当a=b时取等号.我们来看下面3个问题:问题1:已知x,y为正数,求（x+y）（1/x+4/y）的最小值.问题2:已知z,y为正数且满足1/x+1/y=2,求x+2y的最小值.     

怎样证明条件等式

有关证明条件等式的代数题,是一类综合性比较强的题目,如果能让学生掌握其各种不同的证明方法,对于培养他们的逻辑思维能力和熟练的技能技巧都是大有益处的。下面介绍几种证明条件等式的常用方法。一、将已知条件直接代入欲证等式例1 已知:x=(a-b)/(a b),y=(b-c)/(b c), z=(c-a)/(c a) 求证:(1 x)(1 y)(1 z) =(1-x)(1-y)(1-z) 证明:∵(1 x)(1 y)(1 z) =(1 (a-b)/(a b))(1 (b-c)/(b c))(1 (c-a)/(c a)) =2a/(a b)·2b/(b c)·2c/(c a) (1-x)(1-y)(1-z) =(1-(a-b)/(a b))(1-(b-c)/(b c))(1-(c-a)/(c a)) =2b/(a b)·2c/(b c)·2a/(c a) ∴ (1 x)(1 y)(1 z)=(1-x)(1-y)(1-z) 二、通过已知条件之间的相互变换,得出求证式。例2.设x=by cz,y=cz ax,z=ax by 试证:(a 1)x=(b 1)y=(c 1)z     

一道柯西不等式背景题的变式探究

题目(2012年高考湖北卷·理6)设口,b,c,x,y,z是正数,且a~2+6~2+c~2=10,x~2+y~2+z~2=40,ax+by+xz=20,则a+b+c/x+y+z=A.1/4.B.1/3 C.1/2 D.3/4以上题目旨在考查柯西不等式、等比性质等基础知识.笔者将其进一步推广得到一般性的变式题1、2(如下),并进行探究.变式1设a,b,c,z,y,z,p,q,r.是正数,且a~2+b~2+c~2=p~2,x~2+y~2+z~2=q~2,ax+by+cz=r~2,     

Einstein''''s Answer

One day when Einstein was 1(?) his way home, a young man stopped him and wished 2(?) a word with him. Einstein granted his request. The young man asked, "How, Mr Einstein, can you 3(?) your fame? " The scientist said, "It seems that you have been thinking of 4(?) famous every day." The young man nodded.     

教育问与答

如何纠正孩子的发音? Q我的孩子4岁4个月,说话时发音不怎么准,有些字词说不清楚。如:f的音她会发成b的音,g的音会发成d的音,还有一些卷舌、平舌的都发不大准。有什么快速、方便的办法来纠正吗?     

教育问与答

又换看护人了Q女儿2岁了,从4个月起白天就由奶奶或姥姥来照看,一般半年替换一次。姥姥前两天刚过来,女儿变得特别黏我,而且本来大小便已经很规律了,现在又开始频繁尿裤及尿床。这种两边交替的看护方式会对她造成不良影响吗?在这种看护方式下,我应该做些什么帮助她适应?     

Answer Preference and Answer Conflict in Children's Completions of Concrete Operational Stories

Abstract The procedures used in presenting many concrete operational problems have two drawbacks: children may fail through misunderstanding the questioner's intention and there is a biasing towards a unidimensional interpretational of the question. Story completion procedures were used here in an effort to circumvent these problems. In the first experiment six and seven year‐old children had to select one of four completions of a story of traditional style and content. Selection of completions that involved the coordination of perceptual dimensions was associated with separately‐assessed ability in conservation and in class inclusion. This was interpreted as showing that success and failure on these standard tasks reflects answer preferences which are not specific to the procedures used in the tasks. The second experiment attempted to equalise the biasing towards the unidimensional answer in conservation tasks. The problem was presented through a narrative in which the answers of a nonconserving and of a conserving protagonist came into symmetrical conflict. Selection of the conservation completion by the four and five year‐old subjects was far more common than in the standard situation, suggesting that when the biasing is removed a more basic form of conservation competence is revealed.     

Question & Answer

Q What does his father like?/What is his father like?/What does his father look like?这三句话的意思有何不同?(滁州欣宜)A这里涉及对like意思的理解:like既可以作动词,表示"喜欢"的意思,也可以作介词,表示"(看起来)像……"的意思。你提到的第一句,like是动词,全句意为"他爸爸喜欢什么?"可以用"He likes     

Question & Answer

Q有这样一道同义句改写题,我不知所措:The minute I sat down in front of the TV,my mom came over.=My mom came overI sat down in front of the TV(.池州舟舟)A这里的the minute/moment...相当于as soon as...,意思为"一……就……"。再如:I need to see her the minute she arrives.=I needto see her as soon as she arrives.所以,横线上应填写as soon as。(思迈)