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1、学习必备欢迎下载基本不等式应用一基本不等式1.(1)若Rba,,则abba222 (2)若Rba,,则222baab(当且仅当ba时取“=”)2.(1)若*,Rba,则abba2 (2)若*,Rba,则abba2(当且仅当ba时取“=”)(3)若*,Rba,则22baab (当且仅当ba时取“=”)3.若0 x,则12xx(当且仅当1x时取“=”);若0 x,则12xx(当且仅当1x时取“=”)若0 x,则11122-2xxxxxx即或 (当且仅当ba时取“=”)3.若0ab,则2abba (当且仅当ba时取“=”)若0ab,则22-2abababbababa即或 (当且仅当ba时取“=”)4
2、.若Rba,,则2)2(222baba(当且仅当ba时取“=”)注:(1)当两个正数的积为定植时,可以求它们的和的最小值,当两个正数的和为定植时,可以求它们的积的最小值,正所谓“积定和最小,和定积最大”(2)求最值的条件“一正,二定,三取等”(3)均值定理在求最值、比较大小、求变量的取值范围、证明不等式、解决实际问题方面有广泛的应用应用一:求最值例 1:求下列函数的值域(1)y3x 212x 2(2)yx1x解:(1)y3x 212x 223x 212x 26 值域为 6,+)(2)当 x0 时,yx1x2x1x2;当 x0 时,yx1x=(x1x)2x1x=2值域为(,22,+)解题技巧:技
3、巧一:凑项例 1:已知54x,求函数14245yxx的最大值。解:因450 x,所以首先要“调整”符号,又1(42)45xx不是常数,所以对42x要进行拆、凑项,5,5404xx,11425434554yxxxx231当且仅当15454xx,即1x时,上式等号成立,故当1x时,max1y。评注:本题需要调整项的符号,又要配凑项的系数,使其积为定值。技巧二:凑系数例 1.当时,求(82)yxx的最大值。学习必备欢迎下载解析:由知,利用基本不等式求最值,必须和为定值或积为定值,此题为两个式子积的形式,但其和不是定值。注意到2(82)8xx为定值,故只需将(82)yxx凑上一个系数即可。当,即 x2
4、 时取等号当 x2 时,(82)yxx的最大值为8。评注:本题无法直接运用基本不等式求解,但凑系数后可得到和为定值,从而可利用基本不等式求最大值。变式:设230 x,求函数)23(4xxy的最大值。解:230 x023x2922322)23(22)23(42xxxxxxy当且仅当,232xx即23,043x时等号成立。技巧三:分离例 3.求2710(1)1xxyxx的值域。解析一:本题看似无法运用基本不等式,不妨将分子配方凑出含有(x1)的项,再将其分离。当,即时,421)591yxx(当且仅当x1 时取“”号)。技巧四:换元解析二:本题看似无法运用基本不等式,可先换元,令t=x1,化简原式在
5、分离求最值。22(1)7(1+10544=5ttttytttt)当,即 t=时,4259ytt(当 t=2 即 x1 时取“”号)。评注:分式函数求最值,通常直接将分子配凑后将式子分开或将分母换元后将式子分开再利用不等式求最值。即化为()(0,0)()Aymg xB ABg x,g(x)恒正或恒负的形式,然后运用基本不等式来求最值。技巧五:注意:在应用最值定理求最值时,若遇等号取不到的情况,应结合函数()af xxx的单调性。例:求函数2254xyx的值域。解:令24(2)xt t,则2254xyx22114(2)4xtttx因10,1ttt,但1tt解得1t不在区间2,,故等号不成立,考虑单
6、调性。因为1ytt在区间1,单调递增,所以在其子区间2,为单调递增函数,故52y。所以,所求函数的值域为5,2。练习求下列函数的最小值,并求取得最小值时,x 的值.文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2
7、Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D
8、5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG
9、9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X
10、1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU
11、3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7
12、C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8学习
13、必备欢迎下载(1)231,(0)xxyxx(2)12,33yxxx (3)12sin,(0,)sinyxxx2已知01x,求函数(1)yxx的最大值.;3203x,求函数(2 3)yxx的最大值.条件求最值1.若实数满足2ba,则ba33的最小值是 .分析:“和”到“积”是一个缩小的过程,而且ba33定值,因此考虑利用均值定理求最小值,解:ba33 和都是正数,ba33632332baba当ba33时等号成立,由2ba及ba33得1ba即当1ba时,ba33的最小值是6变式:若44loglog2xy,求11xy的最小值.并求 x,y 的值技巧六:整体代换:多次连用最值定理求最值时,要注意取等号
14、的条件的一致性,否则就会出错。2:已知0,0 xy,且191xy,求xy的最小值。错解:0,0 xy,且191xy,1992212xyxyxyxyxy故min12xy。错因:解法中两次连用基本不等式,在2xyxy等号成立条件是xy,在1992xyx y等号成立条件是19xy即9yx,取等号的条件的不一致,产生错误。因此,在利用基本不等式处理问题时,列出等号成立条件是解题的必要步骤,而且是检验转换是否有误的一种方法。正解:190,0,1xyxy,1991061016yxxyxyxyxy当且仅当9yxxy时,上式等号成立,又191xy,可得4,12xy时,min16xy。变式:(1)若Ryx,且1
15、2yx,求yx11的最小值(2)已知Ryxba,且1ybxa,求yx的最小值技巧七、已知 x,y 为正实数,且x 2y 221,求 x1y2的最大值.分析:因条件和结论分别是二次和一次,故采用公式aba 2b 22。同时还应化简1y2中 y2前面的系数为12,x1y2x21y 222 x12y 22文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y
16、9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8
17、ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10
18、V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8
19、文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:C
20、N2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R1
21、0D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4
22、HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8学习必备欢迎下载下面将 x,12y 22分别看成两个因式:x12y 22x 2(12y 22)22x 2y 2212234即 x1y22 x12y 22342 技巧八:已知a,b 为正实数,2baba30,求函数y1ab的最小值.分析:这是一个二元函数的最值问题,通常有两个途径,一是通过消元,转化为一元函数问题,再用单调性或基本不等式求解,对本题来说,这种途径是可行的;二是直接用基本不等式,对本题来说,因已知条件中既有和的形式,又有积的形式,不能一步到位求出最值,
23、考虑用基本不等式放缩后,再通过解不等式的途径进行。法一:a302bb1,ab302bb1b2 b230bb1由 a0 得,0b15 令 tb+1,1t16,ab2t234t31t 2(t16t)34t16t2t16t8 ab18 y118当且仅当 t4,即 b3,a6 时,等号成立。法二:由已知得:30 aba2b a2b22 ab 30ab22 ab令 uab则 u222 u300,52 u32 ab32,ab18,y118点评:本题考查不等式abba2)(Rba,的应用、不等式的解法及运算能力;如何由已知不等式230abab)(Rba,出 发 求 得ab的 范 围,关 键 是 寻 找 到a
24、bba与之 间 的 关 系,由 此 想 到 不 等 式abba2)(Rba,,这样将已知条件转换为含ab的不等式,进而解得ab的范围.变式:1.已知 a0,b0,ab(ab)1,求 ab 的最小值。2.若直角三角形周长为1,求它的面积最大值。技巧九、取平方5、已知 x,y为正实数,3x2y10,求函数 W3x 2y 的最值.解法一:若利用算术平均与平方平均之间的不等关系,ab2a 2b 22,本题很简单3x 2y2(3x)2(2y)22 3x2y 25 解法二:条件与结论均为和的形式,设法直接用基本不等式,应通过平方化函数式为积的形式,再向“和为定值”条件靠拢。W0,W23x2y23x 2y
25、1023x 2y 10(3x)2(2y)2 10(3x2y)20 W20 25 变式:求函数152152()22yxxx的最大值。解析:注意到21x与52x的和为定值。22(2152)42(21)(52)4(21)(52)8yxxxxxx又0y,所以022y当且仅当21x=52x,即32x时取等号。故max2 2y。评注:本题将解析式两边平方构造出“和为定值”,为利用基本不等式创造了条件。总之,我们利用基本不等式求最值时,一定要注意“一正二定三相等”,同时还要注意一些变形技巧,积极创造条件利用基本不等式。应用二:利用基本不等式证明不等式1已知cba,为两两不相等的实数,求证:cabcabcba
26、2221)正数 a,b,c 满足 abc1,求证:(1a)(1b)(1c)8abc文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN
27、2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10
28、D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 H
29、G9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9
30、X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 Z
31、U3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V
32、7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8学习必备欢迎下载例 6:已知 a、b、cR,且1abc。求证:1111118abc分析
33、:不等式右边数字8,使我们联想到左边因式分别使用基本不等式可得三个“2”连乘,又1121abcbcaaaa,可由此变形入手。解:a、b、cR,1abc。1121abcbcaaaa。同理121acbb,121abcc。上述三个不等式两边均为正,分别相乘,得1112221118bcacababcabc。当且仅当13abc时取等号。应用三:基本不等式与恒成立问题例:已知0,0 xy且191xy,求使不等式xym恒成立的实数m的取值范围。解:令,0,0,xyk xy191xy,991.xyxykxky1091yxkkxky10312kk。16k,,16m应用四:均值定理在比较大小中的应用:例:若)2l
34、g(),lg(lg21,lglg,1baRbaQbaPba,则RQP,的大小关系是 .分析:1ba0lg,0lgba21Q(pbabalglg)lglgQababbaRlg21lg)2lg(RQP。文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3
35、V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C
36、6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编
37、码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q
38、3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5
39、J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9
40、R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8文档编码:CN2Q3R10D5J4 HG9R1Y9X1X8 ZU3V10V7C6G8