Nối tiếp nội dung phần 1 Thiết kế bài giảng Giải tích 12 (Tập 1), phần 2 giới thiệu cách thiết kế các bài giảng về hàm số lũy thừa - Hàm số mũ và hàm số lôgarit. Mời các bạn cùng tham khảo nội dung chi tiết.
ChiroNq II HAM SO LUY THUA - HAM SO MU VA HAM SO LOGARIT Phan OTTOlVG VAN D E C U A C H C W ^ G I NOI DUNG Ndi dung chfnh ciia chuong II : ' Luy thira la gi ? Luy thiia vdi sd mu nguyen; Phuang trinh x" = b ; Can bac n ciia mot sd duong; Liiy thira vdi sd mu hiin ti va luy thira vdi sd mu vd ti Cac tfnh chai ciia liiy thira vdi sd mu thuc; Luy thira ciia mdt tfch va mot thuang, tfch hai liiy thira va thuang hai liiy thiia Ham sd y = x" , dao ham ciia ham so y = x**, khao sat ham so y = x" • Logarit la gi? Cac tfnh chat cua Idgarit Mot sd quy tic tfnh Idgarit, Idgarit ciia mot luy thiia, phuang phap ddi co so, Idgarit tu nhien va Idgarit thap phan Ham sd mii va ham sd Idgarit : Khai niem ham sd mu, khao sat ham so mu Khai niem ham sd Idgarit, khao sat ham sd Idgarit Phuong trinh mu va phuong trinh Idgarit : phuong phap giai va mdt so phuong trinh don gian Bai phuang trinh mil va bai phuang trinh logarit : phuang phap giai va mdt sd phuong trinh don gian 192 IL MUC TIEU Kien thiic Nim dugc toan bd kie'n thiic co ban chuang da neu tren, cu the': Khai niem luy thiia vdi sd mu thuc va mdt sd tfnh chat cua nd Ham sd mu la gi; khao sat dugc ham sd mu Khai niem ham sd logarit va khao sat chiing • Giai dugc mdt sd phuang trinh va bai phuong trinh mu, Idgarit Mdi quan he giiia ham sd mii va ham sd Idgarit KT nang Khao sat tdt ham sd mii va ham sd Idgarit • Giai thao phuong trinh, bit phuong trinh mQ va Idgarit - Ve dugc dd thj cac ham sd mii va ham so Idgarit Mdi quan he giiia hai ham sd tren Thai Tu giac, tfch cue, dgc lap va chii ddng phat hien cung nhu ITnh hdi kie'n thiic qua trinh boat ddng - Cam nhan dugc su cin thie't cua ham so mii va ham sd Idgarit thuc te' • Cam nhan dugc thuc te' ciia toan hgc, nha't la dd'i vdi mu va Idgarit Giii tich 12/1 193 Phan Z C A c B A I SOAI!^ §1 Luy thufa (tieTt 1, 2, 3) I M U C T I E U Kien thirc HS nim dugc : Nhd lai liiy thira vdi sd mii nguyen ' xay dung dugc khai niem luy thira vdi so mu thuc Hie'u va van dung dugc mdt so tfnh chat ciia liiy thiia vdi so mu thuc KT nang Sau hgc xong bai nay, HS phai bie't khai niem ciia liiy thira vdi so' mil thuc Van dung dugc cac tfnh chat giai toan Nim dugc mdi quan he giiia liiy thira vdi so mu thuc vdi phuang trinh x"=b " Lien he vdi mdt sd liiy thiia da hgc Thai * Tu giac, tfch cue hgc tap Biei phan biet rd cac khai niem co ban va van dung tirng trudng hgp cu the Tu eac va'n de ciia toan hgc mdt each Idgic va he thdng II CHUAN BI CUA G V VA HS Chuan hi cua GV Chuan bj cac cau hdi ggi md 194 • Chuin bj cac hinh tir hinh 26 den hinh 27 Chuan bj pha'n mau, va mdt sd dd diing khac Chuan bi cua HS Can dn lai mdt sd kien thiic da hgc ve luy thira da hgc d Idp dudi HI PHAN PHOI T H d l LUONG Bai chia lam tiet : Tie't I : Td ddu din hit muc phdn I Tie't : Tie'p theo din hit phdn I Tii't : Tiip theo din hit phdn II IV TIEN TRINH DAY - HOC A OAT VAIN DE Cau hdi Xet tfnh diing - sai cua cac cau sau day : a) Vdi mgi a thi a.a = a^ b) Chi cd a > thi mdi xay a.a = a^ GV : Khing djnh a) diing, cdn khang djnh b) sai Cd the din cac vf du cu the' Cau hdi Thuc hien cac phep tfnh sau : :(uf a) v^y b) [yfnf GV : Sau ddy, chung ta se nghien cdu ve luy thda vdi sd md thuc 195 B BAI Mdl I KHAI NIEM LtJY THlTA HOATDQNGl Luy thira vdi sd mu nguyen • Thuc hien Qy I 5' Hoat ddng ciia GV Cau hdi Hoat dgng cua HS Ggi y tra loi cau hoi Tfnh 1,5"^ 1,5^ =5,0625 G\: ggi HS thue hien HS cd the sii dung may tfnh dien tii bam: 1.5M = 5.0625 Cau hdi Ggi y tra Idi cau hoi r 2^ Tfnh : 3j I 3j 21' HS cd the sir dung may tinh dien tir bam : (2 +3)^3 = Sau dd ta in not a Cau hdi ta dugc ke't qua Ggi y tra loi cau hdi [^=^S Tfnh (Vs)^ HS cd the sii dung may tfnh dien td ba'm: (^^3)^5 = ' • GV neu djnh nghia: Cho n Id mgt sd nguyen duang Vdi a la sdthuc y, luy thica bdc n ciia a la tich cua n thda sda 196 a"' = a.a a n thijfa so Vdi a^O aO=l a - — a" Trong bieu thiic a", ta ggi a la co sd, n la sdmu • GV neu chii y : > Chu y va 0~" khdng cd nghTa HI Neu mdt sd vi du ve luy thiia vdi so mii nguyen f n'\'^ / 3\5 H2 Tfnh V "+ y • GV neu vf du GV ed the lay vf du tuong tu Hoat ddng cua GV Cau hdi Hoat ddng cua HS Ggi y tra loi cau hdi Hay doi so' hang thii nhat ca so Tacd Cau hdi (1 v'° 27"^=3^°.3"^=3 Ggi y tra loi cau hdi Hay doi so hang thir hai ca Tacd (0,2)"^.25"2=5^5~^ = sd Cau hdi Ggi y tra loi cau hdi Hay doi so hang thii hai ca so Tacd—.2^=2-12^=4 128 197 • GV neu vf du GV cd the la'y vf du tuong tu Hoat dgng ciia GV Hoat ddng ciia HS Cau hoi Ggi y tra loi cau hdi Hay phan tich ngoac [aV2(l + a^) - 2V2a] nhan tir = (aV2+ a ^ V - a V ) = aV2(a^ - ) Cau hdi Ggi y tra loi cau hoi Hay phan tich ngoai ngoac nhan tur Cau hdi Tfnh B 1 a^il-a"^) a^-a aia^ 1) Ggi y tra loi cau hoi B= V2 HOAT DONG 2 Phirong trinh x" = b • Thuc Men ^ 5' Hoat ddng cua GV Cau hdi Hoat dgng cua HS Ggi y tra loi cau hdi Hay bien luan so nghiem • Vdi mgi e R, phuong trinh o phuang trinh x = b X = b ludn cd mdt nghiem Can hdi Ggi y tra loi cau hoi Hay bien luan sd nghiem Vdi < 0, phuang trinh x^ = b phuang trinh x"^ = b khong cd nghiem 198 • Vdi = 0, phuong trinh x = cd mdt nghiem x = • Vdi > 0, phuang trinh x = cd hai nghiem trai da'u • GV dua nhan xet: Dd thi cua hdm sd y = x^*"*"' tuang tU dd thi hdm sd y = x^ vd dd thi hdm sd y = x'^* tuang tu thi hdm sd y = x'^ Tu: dd ta cd ke't qud bien ludn so nghiem cua phuang trinh x" = b nhU sau Trudng hap n le Vdi mgi sdthUc b, phuang trinh cd nghiem nhdt Trudng hap n chdn Vdi b < 0, phuang trinh vd nghiem ; Vdi b = 0, phuang trinh cd mgt nghiem x = ; Vdi b > phuang trinh cd hai nghiem trdi dd'u H3 Tim sd nghiem phuang trinh : \^ = 2008 , x^°°^ = -2008 H4 Tim sd nghiem phuang trinh : x^°°^ = -2009, x^°°^ = va x^^^^ = 2009 HOAT DONG 3 Can bac n • GV neu van de : Cho sd nguyen duang n, phuang trinh a"" =b dua de'n hai bdi todn ngugc : •Bie't a tinh b •Bie't b tinh a Bdi todn thd nhdt la tinh luy thica cua mgt sd Bdi todn thic hai ddn de'n khdi niem lay cdn cua mdt sd 199 • GV neu djnh nghTa : Cho sdthuc b vd sd nguyen duang n >2 Sda dugc ggi la cdn bdc n cua sdb ne'u a " = b H5 Hay chiing minh la can bac hai cua H6 -2 cd phai la can bac hai cua hay khdng? • GV neu bien luan: Tddinh nghia vd kit qud bien lugn ve sd nghiem cua phuang trinh x'' — b, ta cd : Vdi n le, Z? € R phuang trinh cd nhdt mgt cdn bdc n ciia b, ki hieu Id'ilb ^ b 0),ta t"^ -6t + = dugc phuang trinh nao theo t? Ggi y tra Idi cau hoi Cau hoi Phuang trinh cd hai nghiem la Hay giai CSLU h vay X = 0, X = cau e Hoat dgng ciia HS Hoat dgng cua GV Cau hoi Chia hai ve cho 16" ta dugc phuang trinh nao? Cau hdi Ggi y tra Idi cau hoi 2x ^3^ 4j '3^ X + , - = Ggi y tra Idi cSu hoi Hay giai cau c Dat *= j ] (vdi ^ > 0), ta dugc 4^2 + i - = Phuang trinh chi cd mdt nghiem duang t = -T- Vayx = caud Hoat dgng cua HS Hoat dgng cua GV cau hoi Ggi y tra Idi cau hoi Tim dieu ki6n xac dinh cua x > l phuang trinh Cau hoi Hay giai cSu d Ggi y tra Idi cau hdi Vdi dieu kien x > 1, ta cd 339 log X > nen phuang trinh da cho trd l o g ( x - l ) = vay X = cau e Hoat ddng cua GV Cau hdi Hoat ddng cua HS Ggi y tra Idi cau hoi Tim dieu kien xac dinh cua x > phuang trinh Ggi y tra Idi cau hdi Cau hdi Dua vi ciing mdt co so, ta dugc Hay giai cau e log3 X + log , X + log^.i X = 32 «> log3 X + log3 X - log3 X = hay log3 x = Vay x = 27 cauf Hoat ddng ciia GV Cau hdi Hoat dgng cua HS Ggi y tra Idi can hdi Tim dieu kien xac dinh cua Dieu kien chung ciia phuang trinh la phuang trinh X > Can hdi Hay giai cau c Ggi y tra Idi cau hoi rj, , Ta CO X + = X x-1 Tir x^ - 2x - = Phuang trinh cd mdt nghiem x = thoa man dilu kien x > 340 Bai Hudng ddn Dua vao cac phuong phap giai bat phuang trinh mu cau a Hoat ddng cua HS Hoat ddng cua GV Cau hdi Ggi y tra Idi cau hdi Hay chuyen cac sd' hang ve HS tu chuyen ciing ca so Ggi y tra loi cau hdi Cau hoi 1.22''+1.22''+-.22"> 448 Hay giai cau a o - ^ ' ' >448 512 = 2^ «^ 2x> 9 vay J C > - caub Hoat ddng ciia HS Hoat ddng ciia GV Cau hdi Ggi y tra Idi cau hoi Hay chuyen cac sd hang v^ Tacd 0,4 = va2,5 = | cimg ca sd' Cau hdi Ggi y tra Idi cau hdi Hay giai cau b f2f Dat ^ = — , vdi ^ > 0, ta dugc , _^ 2'7^ ' ^^ 2^^ - 3^ - > (^ > 0) Nghiem cua bat phuang trinh la ^ > — Vay X < - 341 cauc Hoat ddng cua HS Hoat ddng cua GV Cau hoi Ggi y tra Idi cau hdi Tun tap xac dinh cua b^t \2_i>0 phuang trinh Tacd • l o g i ( x - l ) > 0 Cau hoi Ggi y tra Idi cau hoi Hay giai cau c log3logi(x^ - l ) < l o g 3 0x2>f « ^^