《高中數(shù)學 第一章 推理與證明 1.2 綜合法與分析法 1.2.1 綜合法課件 北師大版選修22》由會員分享，可在線閱讀，更多相關《高中數(shù)學 第一章 推理與證明 1.2 綜合法與分析法 1.2.1 綜合法課件 北師大版選修22（25頁珍藏版）》請在裝配圖網(wǎng)上搜索。

1、2 2綜合法與分析法綜合法與分析法2 2.1 1綜合法MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識

2、梳理了解綜合法的思考過程,會用綜合法證明一些數(shù)學問題. MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知

3、識梳理綜合法從命題的條件出發(fā),利用定義、公理、定理及運算法則,通過演繹推理,一步一步地接近要證明的結(jié)論,直到完成命題的證明.我們把這樣的思維方法稱為綜合法.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳

4、理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITA

5、NGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHI

6、SHISHULI知識梳理題型一題型二題型三MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題

7、型二題型三反思反思此題用綜合法證明時,可以先從條件出發(fā),也可以先從基本不等式出發(fā),通過換元、拼湊等方法構(gòu)造定值.若連續(xù)兩次或兩次以上利用基本不等式,則需要注意這幾次利用基本不等式時等號成立的條件是否相同.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGY

8、ANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI

9、SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIA

10、NLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三【例2】如圖,正三棱柱ABC-A1B1C1的棱長均為a,D,E分別為C1C與AB的中點,A1B交AB1于點G.求證:(1)A1BAD;(2)CE平面AB1D.分析:(1)為了證明A1BAD,可證A1B平面AB1D,連接DG,顯然A1BAB1,所以證明A1BDG,可利用A1DB是等腰三角形以及點G是A1B的中點得證.(2)要證CE平面AB1D,只需證CE與平面AB1D內(nèi)的一條直線(DG)平行即可.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN

11、隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三證明:(1)連接A1D,DG,BD.如圖,三棱柱ABC-A1B1C1是棱長均為a的正三棱柱,四邊形A1ABB1為正方形,A1B

12、AB1.D是C1C的中點,A1C1D BCD.A1D=BD.易知G為A1B的中點,A1BDG.又DGAB1=G,A1B平面AB1D.AD平面AB1D,A1BAD.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI

13、知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三(2)連接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.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHU

14、LI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三【變式訓練2】 如圖,在直三棱柱ABC-A1B1C1中,A1B1=A1C1,D,E分別是棱BC,CC1上的點(點D不同于點C),且ADDE,F為B1C1的中點.求證:(1)平面ADE平面BCC1B1;(2)直線A1F平面ADE.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理

15、目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三證明:(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=

16、E,AD平面BCC1B1.又AD平面ADE,平面ADE平面BCC1B1.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHIS

17、HISHULI知識梳理題型一題型二題型三(2)A1B1=A1C1,F為B1C1的中點,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.由(1)知AD平面BCC1B1,故A1FAD.又AD平面ADE,A1F平面ADE,A1F平面ADE.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳

18、理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI

19、典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLI

20、AN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二題型三【變式訓練3】 已知數(shù)列an滿足a1=1,a2=3,an+2=3an+1-2an(nN+).(1)證明:數(shù)列an+1-an是等比數(shù)列;(2)求數(shù)列an的通項公式.(1)證明:因為an+2=3an+1-2an,所以an+2-an+1=2an+1-2an=2(an+1-an),所以 .又a2-a1=2,所以數(shù)列an+1-an是以2為首

21、項,2為公比的等比數(shù)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+).MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYAN

22、LIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 6A.-2B.0C.1D.2答案:CMUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANG

23、YANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 62已知角A,B為ABC的內(nèi)角,則“AB”是“sin Asin B”的()A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件角A,B為ABC的內(nèi)角,sin A0,sin B0.sin Asin B2Rsin A2Rsin BabAB(其中R是ABC外接圓的半徑).答案:CMUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識

24、梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 6MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITO

25、UXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 6MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANG

26、YANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 65若sin +sin +sin =0,cos +cos +cos =0,則cos(-)=. 解析:因為已知條件中有三個角,而所求結(jié)論中只有兩個角,所以我們只需將已知條件中的角消去即可,依據(jù)sin2+cos2=1消去

27、.由已知,得sin =-(sin +sin ),cos =-(cos +cos ),則(sin +sin )2+(cos +cos )2=sin2+cos2=1,MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 6