(完整word版)高考数学第一轮复习教案-基本不等式(word文档良心出品).pdf
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1、高三数学一轮复习 10.4 基本不等式一、课标要求:1.解基本不等式及成立条件.2.能应用基本不等式判断大小求最值.3.应用基本不等式解决实际问题和综合问题.二、重难点:1.重点:正确应用基本不等式进行判断和计算.2.难点:基本不等式的变形应用.三、教学方法:以启发引导,探索发现为主导.讲解练习为主线.用一题多解,一题多变突出重点,突破难点.以综合应用提高分析解决问题的能力,培养创新能力.四、教学过程:(一)、学情评估,导入新课:1.下列不等式中不一定成立的是()A.222ababB.222()ababC.12aaD.2212aa2.0,0,2mnmn,则mn的最大值为。3.0,0 xy,且1
2、91xy,则xy最小值是。(二)、探求、归纳知识体系:1.基本不等式:222abab(,a bRxy)2abab(0,0)ab2baab(0)ab变形:222()22ababab2222ababab(,)a bR2.基本不等式与最值:若,x yR和定积最大:若xys,则24sxy(当且仅当xy时“=”成立)积定和最小:若xyp,则2xyp(当且仅当xy时“=”成立)注意一:要用此结论需满足三个条件:简称:一正二定三相等注意二:条件不足时可通过拆分与配凑创设条件。(三)基本不等式的应用:例一:设0,0 xy,且440 xy,求lglgxy的最值变式训练.若221xy,求(1)(1)xyxy的最小
3、值。(变形应用).函数214yxx的最大值为。例二:若0 x,求12()3f xxx的最小值。若0 x,求12()3f xxx的最大值。归纳:1(0)yxxx的值域是什么?变式训练二:求4()3lglgf xxx,(1)x的最小值。(变形应用)求14245yxx,5()4x的最小值。(对比应用)若12x,则1xx的最大值为。文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6
4、H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8
5、 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M
6、6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V
7、6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F
8、9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M
9、5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:
10、CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5例三:(能力提高)若正数,x y满足21xy求11xy的最小值。变式训练三.已知0,0 xy,且191xy,则xy最小值为()A.12 B.16 C.6 D.24 例四某商品进货价为每件50 元。据市场调查,当售价(每件x元)在5080 x时,每天售出的件数5210(40)px,若想每天获利最多,价格应定为每件多少元?例五(反思)辨析正误,错的说出原因。求225()2loglogxxf x(01x)的最值。解:225()22 log22
11、5logxxf x求224()sinsinf xxx的最小值。解:222244()sin2 sinsinsinf xxxxx4四:课堂小结:这一节课的收获是:文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6
12、H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8 HG10M6S9A5V6 ZT3F9H7M4M5文档编码:CN10I6H8R7P8
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