北师大版高中数学选修2 2同步配套课件：12 综合法与分析法121.pptx

1、-1-2-ZHISHI SHULI知识梳理SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航了解综合法的思考过程,会用综合法证明一些数学问题.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航综合法从命题的条件出发,利用定义、公理、定理及运算法则,通过演绎推理,一步一步地接近要证明的结论,直到完成命题的证明.我们把这样的思维方法称为综合法.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理SUITANGYA

2、NLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三反思反思此题用综合法证明时,可以先从条件出发,也可以先从基本不等式出发,通过换元、拼凑等方法构造定值.若连续两次或两次以上利用基本不等式,则需要注意这几次利用基本不等式时等号成立的条件是否相同.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI

3、典例透析目标导航题型一题型二题型三ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三【例2】如图,正三棱柱ABC-A1B1C1的棱长均为a,D,E分别为C1C与AB的中点,A1B交AB1于点G.求证:(1)A1BAD;(2)CE平面AB1D.分析:(1)为了证明A1BAD,可证A1B平面AB1D,连接DG,显然A1BAB1,所以证明A1BDG,可利用A1DB是等腰三角形以及点G是A1B的中点得证.(2)

4、要证CE平面AB1D,只需证CE与平面AB1D内的一条直线(DG)平行即可.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三证明:(1)连接A1D,DG,BD.如图,三棱柱ABC-A1B1C1是棱长均为a的正三棱柱,四边形A1ABB1为正方形,A1BAB1.D是C1C的中点,A1C1D BCD.A1D=BD.易知G为A1B的中点,A1BDG.又DGAB1=G,A1B平面AB1D.AD平面AB1D,A1BAD.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型

5、二题型三(2)连接GE,GE A1A,GE平面ABC.DC平面ABC,GEDC.ECGD.又EC平面AB1D,DG平面AB1D,EC平面AB1D.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三【变式训练2】如图,在直三棱柱ABC-A1B1C1中,A1B1=A1C1,D,E分别是棱BC,CC1上的点(点D不同于点C),且ADDE,F为B1C1的中点.求证:(1)平面ADE平面BCC1B1;(2)直线A1F平面ADE.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题

6、型一题型二题型三证明:(1)三棱柱ABC-A1B1C1是直三棱柱,CC1平面ABC.AD平面ABC,CC1AD.又ADDE,CC1平面BCC1B1,DE平面BCC1B1,CC1DE=E,AD平面BCC1B1.又AD平面ADE,平面ADE平面BCC1B1.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三(2)A1B1=A1C1,F为B1C1的中点,A1FB1C1.CC1平面A1B1C1,且A1F平面A1B1C1,CC1A1F.又CC1平面BCC1B1,B1C1平面BCC1B1,CC1B1C1=C1,A1F平面BCC1B1.

7、由(1)知AD平面BCC1B1,故A1FAD.又AD平面ADE,A1F平面ADE,A1F平面ADE.ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三【变式训练3】已知数列an满足a1=1,a2=3,an+2=3an+1-2an(nN+).(1)证明:数列an+1-an是等比数列;(

8、2)求数列an的通项公式.(1)证明:因为an+2=3an+1-2an,所以an+2-an+1=2an+1-2an=2(an+1-an),所以 .又a2-a1=2,所以数列an+1-an是以2为首项,2为公比的等比数列.(2)解:由(1)得an+1-an=2n(nN+).所以an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=2n-1+2n-2+2+1=2n-1(nN+).ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 5 6A.-2B.0C.1D.2答案:CZHISHISHULI知识梳理SUITANG

9、YANLIAN随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 5 62已知角A,B为ABC的内角,则“AB”是“sin Asin B”的()A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件角A,B为ABC的内角,sin A0,sin B0.sin Asin B2Rsin A2Rsin BabAB(其中R是ABC外接圆的半径).答案:CZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 5 6ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标

10、导航1 2 3 4 5 6ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 5 65若sin+sin+sin=0,cos+cos+cos=0,则cos(-)=.解析:因为已知条件中有三个角,而所求结论中只有两个角,所以我们只需将已知条件中的角消去即可,依据sin2+cos2=1消去.由已知,得sin=-(sin+sin),cos=-(cos+cos),则(sin+sin)2+(cos+cos)2=sin2+cos2=1,ZHISHISHULI知识梳理SUITANGYANLIAN随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 5 6