Phương pháp và kĩ thuật ôn nhanh thi đại học đạt điểm cao môn Toán cuốn sách hay dùng để ôn tập những kiến thức môn toán để đạt điểm cao. Bên cạnh đó còn hướng dẫn các em cách giải bài tập toán nhanh và hiệu quả, nhằm mục đích giúp các em đạt điểm cao nhất trong các bài kiểm tra, kì thi sắp tới.
510.76 PH561P n c u Y i n P I I U Kiirinii PHUONO PHAP on niinnii TIII oni HOC D A T DIEM CAO NGUYEN PHU KHANH PHUONG PHAP & KI THUAT on niinnii TIII oni HOC DAT DIEM CAO \SP/ NHA XUAT BAN DAI HOC SlT PHAM Phucmg phdp vd ki thudt on nhanh thi Dai hoc dat diem cao mon Todn giiip hoc sinh on tap, van dung sang tao kien thuc, tong hop, phan tich, kiem tra nang lye toan dien, hucfng dan chien thuat giai de thi dai hoc Sach trinh bay noi dung tam on thi dai hoc gbm: - Phucmg trinh, bat phuong trinh, he phucmg trinh dai so He phucmg trinh mil va Idgarit - Bai todn to'ng hap Bat dang thuc; cue tri cua bieu thuc dai so - phuang phdp giai phuang trinh lugng gidc nhanh nhat - Van de lien quan den sophiec, dai sotohffp, xdc sudi - Cdc bai todn lien quan den ung dung ciia dqo ham va thi cua ham so: chieu bieh thien cua ham so; cue tri; gid tri Ion nhat va nho nhat cua ham so; tiep tuyen, tiem can (dimg vd ngang) cua thi ham so; tint tren thi nhirng diem c6 tinh chat cho truac, luang giao giua hai thi (mot hai thi Id dubng thang) - Tim giai han - Tim nguyen ham, tinh tich phan Ung dung cua tich phan: Tinh dien tich hinh phang, the tich khdi trdn xoay - Hinh hoc khong gian (tong hap): quan he song song, quan he vuong goc cua dubng thang, mat phdng; dien tich xung quanh cua hinh non trdn xoay, hinh tru trdn xoay; the tich khdi lang tru, khdi chop, khdi non trdn xoay, khdi tru trdn xoay; tinh dien tich mat can vd the tich khdi cdu - Phuang phdp toa mat phang: Xdc dinh toa cua diem, vecta Duong trdn, ba dudng conic Viet phuang trinh dudng thdng Tinh gdc; tinh khodng cdch tit diem den dudng thang - Phuang phdp toa khdng gian: Xdc dinh toa cua diem, vecta Duong trdn, milt cdu Viet phuang trinh mat phdng, dudng thdng Tinh goc; tinh khoang cdch tie diem den dudng thang, mat phdng; khodng cdch giita hai dubng thdng; vi tri tuang ddi cua dubng thdng, mat phdng vd mat cdu Mac du tac gia da danh nhieu tam huyet cho cuon sach, nhung sai sot la dieu kho tranh khoi, rat mong nhan duac sy phan bien va gop y quy bau ciia ban doc de nhung Ian tai ban sau cuon sach dugc hoan thien hon Tac gia CAU T R U C D E T H I DAi HQC M O N TOAN' I P H A N C H U N G (7 d i i m ) Cau (2 diem): a) Khao sat su bien thien va ve thi cua ham so b) Cac bai toan lien quan den u n g dung cua dao ham va thi cua ham so: chieu bien thien ciia ham so; cue t r i ; gia trj idn nhat va nho nhat cua ham so; tiep tuyen, tiem can (dung va ngang) cua thi ham so; t i m tren thi n h i i n g diem c6 tinh chat cho truac, tuong giao giira hai thi (mot hai thi la d u o n g thang) Cau (1 diem): Cong thiic lugng giac, phuong trinh lugng giac Cau (1 di£m): Phuong trinh, bat p h u o n g trinh, he phuong trinh dai so Cau (1 diem): - T i m gioi han - Tim nguyen ham, tinh ti'ch phan - u n g dung ciia tich phan: tinh dien tich hinh phang, the tich khoi tron xoay Cau (1 d i i m ) : Hinh hoc khong gian (tong hop): quan he song song, quan he vuong goc cua d u o n g thang, mat phang; dien tich xung quanh cua hinh non tron xoay, h i n h t r u tron xoay; the ti'ch khoi lang t r u , khoi chop, khoi non tron xoay, khoi tru tron xoay; tinh dien tich mat cau va the tich khoi cau Cau (1 diem): Bai toan tong hop I I P H A N R I E N G (3 diem) Thi sink chi dugc lam mot hai phan (phan A hoac phan B) A Theo chucmg trinh Chuan: Cau 7a (1 diem): Phuong phap toa mat phang: - Xac d i n h tpa dp cua diem, vecto - Duong tron, elip - Viet phuong trinh d u o n g thSng - Tinh goc; tinh khoang each t u diem den d u o n g t h i n g Cau 8a (1 d i i m ) : Phuong phap toa khong gian: - Xac d j n h toa dp cua diem, vecto - Duong tron, mat cau - Tinh goc; tinh khoang each t u diem den d u o n g thang, mat phang; khoang each giiia hai duong thang; v i t r i tuong doi cua duong thang, mat phang va mat cau Cau 9a (1 diem): - So phuc - To hop, xac suat, thong ke - Bat dang thiic; eye t r i cua bieu thiic dai so B Theo chuang trinh Nang cao: C a u 7b (1 d i i m ) : Phuong phap tpa mat phSng: - Xac d i n h toa cua diem, vecto - D u o n g tron, ba d u o n g conic - Viet phuong trinh d u o n g thang - Tinh goc; tinh khoang each t u diem den d u o n g th5ng C a u 8b (1 d i i m ) : Phuong phap tpa khong gian: - Xac dinh tpa dp cua diem, vecto - D u o n g tron, mat cau - Viet phuong trinh mat phang, d u o n g thang - Tinh goc; tinh khoang each t u diem den duong thSng, mat p h i n g ; khoang each giua hai duong thang; v i t r i tuong d o i cua d u o n g thang, mat phang va mat cau C a u 9b (1 diem): - So phuc - Do thi ham phan thue h i i u t i dang y = + bx + e px + q quan - Su tiep xiie cua hai d u o n g cong - He p h u o n g trinh m u va iogarit - To hop, xac suat, thong ke - Bat dang thiie Cue t r i eiia bieu thiic dai so PHlTdNG TRINH, BAT PHlTdNG TRINH, HE I PHlTOfNG TRINH Phuong trinh, bat phuong trinh, he phuong trinh dai so H( phuong trinh mil va Iogarit M O T SO H E P H U O N G T R I N H C O BAN 1) He bac nhat hai an, ba an 2) He gbm mot phuong trinh bac nhat va phuong trinh bac cao He ba p h u o n g trinh - He hai p h u o n g trinh - Phuong phap ehung: Su d u n g p h u o n g phap the • 3) He doi xung loai Phuong phap chung: Dat an phu a = x + y; b = xy 4) He doi xung loai Phuong phap chung: T r u tung vehai phuong trinh da cho ta dugc: (x - y).f(x; y) = 5) He phuong trinh dang cap bac hai Xet truong hop y = V o l y ?t 0, ta c6 the tien hanh theo cac each sau: - Dat an p h u y = t.x - Chia ca hai ve'cho y ^ , va dat t = — y MOT SO P H U O N G PHAP G I A I H E P H U O N G T R I N H 1) Phuong phap the • Phuong phap: Ta riit mot an (hay mot bieu thuc) t u mot p h u o n g trinh he va the vao phuong trinh lai • Nhqn dang: Phuong phap thuong hay su d u n g k h i he c6 mot phuong trinh la bac nha't doi voi mot an nao 2) Phuong phap cong dai so 3) Phuong phap bien doi tich 4) Phuong phap dat an p h u 5) Phuong phap ham so 6) Phuong phap su d u n g bat dang thuc C/iM y: ting dung dao ham gidi todn phuong trinh, he phuong trinh Su dung cac tinh chat ciia ham so de giai phuong trinh la dang toan kha quen thupc Ta thuong c6 ba huong ap dung sau day: Huang 1: Thuc hien theo cac buoc: Buac 1: Chuyen phuong t r i n h ve dang: f(x) = k Buac 2: Xet ham so y = f ( x ) Buac 3: Nhan xet: • V o i x = Xg f(x) - f(xg) = k, do X g la nghiem • Voi X > Xg o f(x) > f(xQ) = k, do phuong trinh v6 nghiem • Voi X < Xg f(x) < f(xQ) = k, do phuong trinh v6 nghiem • Vay X g la nghiem nhat a i a phuong trinh Huong 2: Thuc hien theo cac buoc: Suae 1: Chuyen phuong trinh ve dang: f(x) = g(x) Buoc 2: Dung lap luan khang d i n h rang f(x) va g(x) c6 n h u n g tinh chat trai ngugc va xac d i n h X g cho f ( X g ) = g(xg) Buac 3: Vay Xg la nghiem d u y nhat ciia phuong trinh Huang 3: Thuc hien theo cac buac: Buac 1: Chuyen p h u o n g trinh ve dang f(u) = f ( v ) Bwac 2: Xet ham so y = f ( x ) , dung lap luan khang d i n h ham so don dieu Buac 3: K h i f(u) = f(v) o u = v Cac vi du| V i du Giai cac p h u o n g trinh sau tren tap so thuc: V x - l - x + 13 ~ + X X V =x+ 2x V +-=-2x-4 + X x ' " ' + ^ ' ' - ' ' ' + = 2^''*'+2''"+9.2^*" X Lcri giai Dieu kien: x > — Phuong trinh da cho tuong d u o n g v o i VSx - - V3x+ 13 = -if VSx - - V3x +13)(VSx - + V3x + ) Truang hap 1: V S x - l - %/3x + 13 = o V s x - l = V s x T l S , binh phuong hai ve roi riit gon ta dugc x = (thoa man) Trucrng hap 2: V S x - l + V3x + 13 = (1) Neu x > l thi V T ( l ) > >/4 + Vl6 = , neu x < l thi VT ( l ) < ^/4 + Vl6 = De thay x = la nghiem p h u o n g trinh (1) Vay p h u o n g trinh ban dau c6 hai nghiem x = , x = 2, Dieu kien: + - ^ o x X > o x < - hoac x > K h i p h u o n g trinh da cho viet lai: x ^ l + - = -2x^ - 4x + + Vdi X > , phuong trinh o > y x + x = - ( x + x ) + Dat: t = V x ^ + x , t > , ta c6 2t^ + - = o t = hoac t = - - , doi chieu dieu kien ta dugc t = l , tuc la phai c6 Vx^ + x = o x^ + 2x = , phuong trinh c6 nghiem x = -l + \f2 thoa man dieu kien + V d i X ^ - , p h u o n g t r i n h o - V x ^ +2x = -2(x^ + 2x) + Dat t = > / x ^ + x , t > , ta c6: 2t^ - - = o t = | hoac t = - l , d o i chieu dieu kien ta dugc t = ^ , tuc la phai c6 4(x^ + 2x) = 4x^ + 8x - = , phuong trinh c6 nghiem x = ~ ^ '^'^''^ man dieu kien Vay phuong trinh da cho c6 nghiem x = -4-V52 ; X = -1 + r/2 3.Dieu kien: x # , x - i > , x - - > X X Phuong trinh da cho dugc bien doi ve dang: x - — = Ix - — - p x - — (1) X V X V X A p dung cong thuc: — - j = , v d i a > 0; b ^ 0; a ^ b va Va + Vb > Va-vb - X Khi ( l ) o x — = X ^ (2) x-UJ2x-^ * Neu X — > thi — x < 0, t u (2) ta c6 ve trai Ion hon 0, ve phai be hon 0, v6 li X x 4 * Neu X — < thi — x > 0, t u (2) ta c6 ve trai be hon 0, ve phai Ion hon 0, v6 li X X Vay X — = o x - = = > x = (thoa man dieu kien) X Phuong trinh da cho c6 nghiem nhat x = Phuong trinh da cho tuong d u o n g v6i (2'" +2 o - ( " +2 (2"+2")'-2 , phuong trinh da cho tro thanh: t^ - 6t^ - 72t +189 = c = > ( t - ) ' ( t ' + t + l ) = o t = 3, v i t ' + t + 21 > , V t > Voi t = o " + - = « " = ^ ^ « x = log/^*^^ / /— 3±V5 Vay p h u o n g trinh da cho c6 nghiem nhat x = log^ V i d u Giai cac phuong trinh sau tren tap so thuc: log25(x^-8x + 15)^ = | l o g ^ ^ + log5|x-5 2 l o g , (4 - x ) ' + ^ log, (x + ) ' = + log, (x + 6y 4" (V2.6"-4" + V4.24" - " ) = 27" -12" + 2.8" LOT gidi Dieu kien: x > 1;x # 3;x # Phuong trinh da cho tuong duong v o i logs x'' - x + 15 = ' g ^ + l0g5 X - X-1 o l o g g x - +log5 X - = l o g - y - + l o g | x - o | x - = x - l < : = > x - = x - l hoac x - = l - x x = - (vi ^ Vay p h u o n g trinh c6 nghiem x = - x^5) ' Dieu kien: - < x < va x ;t - (*) I^huong trinh da cho o g ^ ( - x ) - o g ^ x + = 3-31og^ {x + 6) o log^ (4 ~ x) + iog^ (x + 6) = + log^ |x + 2| o (4 - x ) ( x + 6) = 4|x + 4(x + 2) = ( - x ) ( x + 6) hoac 4(x + 2) = - ( - x ) ( x + 6) (vi (*) nen ( - x ) ( x + ) > ) + V6i x - + x - = 0c:>x = 2,x = - + V6i x ' - x - = 0x = l + 733,x = l - V 3 Vay p h u o n g trinh c6 hai nghiem x = 2; x = - V33 Dieu kien: 2.6" - 4" > va 4.24" - 3.16" > « x > log, - Ta C O cac danh gia sau: 4"^2.6" - " = 2".2"^2.6" - " < ^ " (4" + 2.6" - " ) = 12" 4" V4.24"-3.16" = 4" ^^4" ( " - " ) < ^ " (2" + V " - " ) = - " + i \ " V " - " < - " + - " ( " +4.6" - " ) = 12" 2 ^ ' Do 4" (V2.6" - " + ^ " - " ) < 2.12", dSng thuc xay k h i x = " - " + " = 27" +8" -12" > 3N/27".8".8" -12" = 2.12", dau bang xay x = Vay p h u o n g t r i n h phai c6 nghiem x = V i d u Giai cac phuong trinh sau tren tap so thuc: x ^ + x ^ + x + l = 3| x + 2 log4 N/X^ - + 3^2 log4(x + 3)^ = 10 + log2(x - 3)^ Lai gidi l D a t 3|^i±?.=y + i c ^ ^ i ± i = y + y + y + i o x = y ^ + y ^ + y (1) 10 T h e o d e b a i ta CO p h u a n g t r i n h x +6x + x + l = y + y = 2x +6x +6x (2), Ttr (1) va (2) s u y 2x^ + 6x2 + 6x - (2y^ + 6y2 + y ) = y o X o 2(x^ - y^) + 6(x2 - y ^ ) + 7(x - y ) = (x - y ) ( x + 2y2 + x y + 6x + y + 7) = o x = y h o a c 2x^ + y + x y + 6x + y + = (3) P h u a n g t r i n h (3) v n g h i e m v i 2x^ +2y^ + x y + 6x + y + = ( x + y)2 + ( x + y ) + + (x + l ) +{y + = (x + y + ) + ( x + l ) + ( y + l ) + l > l lf +1 Vx,y ^i-#-Vay x = y , t h a y v a o (2) ta d u o c 2x'' + 6x2 + 6x = x o + 6x2 + 5x = o ^^2x2 + 6x + 5) = C5> x = V a y X = la n g h i e m p h u o n g t r i n h D i e u k i e n : x < - hoac x > 81og4 7x2 - + j o g ( x + 3)2 = 10 + \og^(x 2log2(x2 -sf - ) + j l o g ( x + 3)2 = 10 + l o g ( x - ) l o g ( x + 3)2 + 3^/log2(x + 3)2 - = D a t t = l o g ( x + 3)2 > Ta CO p h u a n g t r i n h : t^O t2+3t-10 = o t = o l o g ( x + 3)2 = o (x + 3)2 = 16 X = - h o a c x D o i c h i e i i d i e u k i e n , ta c6 n g h i e m p h u a n g t r i n h la x = - V i d u G i a i cac p h u a n g t r i n h sau t r e n t a p so t h u c : x^ + 3x2 + 9x + + ( x - i o ) V - x = V x ^ + yjx^ +x^ +x + \ + Vx'* - Lai giai D i e u k i e n : x < Bien d o i p h u a n g t r i n h ve: (x +1)"' + 6(x +1) = V - x ( - x + 6) Dat u = x + l , v = \ / - x ^ T a c p h u a n g t r i n h : u ' ' + u = v'' + 6v o ( u ' ' - v^) + ( u - v ) = u = V hoac u2 + u v + v2 + = Trudmg hQfp 1: u2 + u v + v2 + = ( u + —)2 + Trumtghgp 2: u = v V - x = x + + = , v6 nghiem x + 1^0 X = -3+N^ - x = x2+2x + l 11 Lot gidi Cdch 1: Goi I la trung diem ciia B C , d u n g A H _L A ' l ( H e A ' l ) , ta c6: BCIAI , , A' BC1(AIA')=:>BC±AH B C l AA' B' AH±BC / AH±(A'BC) A H l A'l AH = d(A,(A'BC)) = a Goi { } = A ' C n A C ' thi O la trung diem cua doan A C , dong thoi O ciing la giao diem ciia \ ~ ~ ~ A C vdi ( A ' B C ) , d(C',(A'BC)) Suy OC , — = =1 d(A,(A'BC)) OA d(c(A'BC)) = d(A,(A'BC))= Ta c6: sin a = ^ aV^ 10-^^3 d(C,(A'BC)) BC BC sina ^15 Dat dai canh cua tam giac ABC la x (x > O) ^ Tarn giac B C C vuong tai B cho C C ' = B C ' - B C ' t —x'(1) Tam giac A ' A I vuong tai A , A H la duong cao cua tam giac nay, suy 1 AH' AI' -+ AA' r 1 aV^r fxV^V V 16 O r= 3a- r+ 3x- ^ 5a- - 4x^ -+- 5a^ X ) 5a" - x " 3a- 3x- (Sa' - 4x-) o 48x'' - 63a-x^ + 15a'' = x ' = a' v x" = 5a^ 16 x = a v x = 4~5 * Truong hgp x = a , t u ' ( l ) => C C = ^ The tich ciia khoi lang tru A B C A ' B ' C : v = s_cc = ^ ^ = i : ^ ABC •^Truong hop x = , t u f l ) => C C = 171 The tich cua khoi lang try A B C A ' B ' C : aV5 a>/l5 4 15a' 256 Cach 2: Goi I , I ' Tan luot la trung diem cua BC va B ' C Dat AB = x, A A ' = y Chon he true toa Oxyz cho: O = I , Ox = I A , Oy = IB, Oz = I I ' Ta c6: l(0;0;0), A IA' = xVs ;0;0 B ; f ; ;0;y , IB = , c 0;-^;0 , A' ;0;y , B C = ( ; - x ; y ) , n , „ , , = I A ' , IB _xy W3 X = D u o n g thang qua I , c6 VTCP l A ' c6 phuong trinh la: y =0 [t^R] z = yt Goi H = h c A ^ ^ , r ^ H { xVs t;0;yt , A H = (t-l);0;yt Ta c6: 3x^ 3x^ A H l A ' ^ A H I A ' = o — ( t - l ) + y-t = Oc:>t = ^ ' ^ 3x^+4y^ AH = -2V3xy^ O x V 3x'+4y' 3x'+4y'^ Do A H I I A ' va A H I I B (do AH.iB = o) nen AH = d(A;(A'BC)), suy ra: AH- = 12xV'+9xV' 3axy a' - , 2 ^ 2 t \ ^f- = o — — = —3a'x'-4x-y^+4a'y = ( l ) (3x^+4y^f 3x^+4y^ X y W Mat khac: • ( B C \ ( A ^ ) = a, sin a = ^ nen: 71 Vl5 c o s ( n , , , , , B C ' ) = cos — a = s m a = ã 10 5a^ 5a^ ôx^+y^=:^=^y^=^-x^ 10 (2) Thay (2) vao ( l ) , ta dugc: o x ' - a ' x ' + a ' = Oc:>x- = a ' hoac x ' = 5a^ 16 172 Truang hop 1: = a" => =^ (x; y ) = a V3 , a'>/3 (dvtt) -x-y = - • V ^ , Trumtghap 2' X' = — 5a' 16 => y ' = 15a' 16 aVs aVis / x => (x; y ) = • VA B C A ' B ' C V3 15%/5 a'(dvtt) -x-y = 256 V i d u Cho h i n h lang try tarn giac deu A B C A ' B ' C Canh day c6 dp dai la a, biet goc giua hai d u o n g t h i n g A B ' va B C la 60° Tinh the tinh ciia khoi lang t r y ABC.A'B'C va khoang each giiia hai d u o n g thSng A B ' va B C theo a Lai gidi Cdch 1: (Ban doc tu ve hinh) De x u l i d u kien goc giira hai d u o n g th3ng A B ' va B C ta se sir d y n g cong thuc AB'.BC COS'(AB";BC') = cos AB';BC') AB' BC Ta chpn he vecto n h u sau: A B = a; A C = b; A A ' = c; AB = A C = De y rang a.b = =a .cos^a^b^ = a.a.cos60" = — D o c vuong v d i hai vecto a,b nen tich v6 huong cua c d o i v o i hai vecto deu bang Ta bieu dien AB',BC theo cac vecto a , b , c D e t h a y A B ' = a + c; B C = - a + b + c AB'.BC = - a - +a.b + c- = - — + c\ A B ' = N/a' + c ^ B C = V a - + b - + c - - a b = Va^T? COSTABVBC^) = COSTAB';BC'] = - - ' V Tu ta c6: V , „ , - = -c = (loai) hoac a- +c2 = AA'.S,,, = a ^ ^ = ^ c2= 2a2=* c =a72 De tim va tinh doan vuong goc chung giua hai duong thang AB' va B C ta lam n h u sau: Gpi cac diem H e A B ' va K e B C cho: A H = x / a ' = x(a + c j ; i3k = y B C = y ^ - a + b + c HK = H A + AB + IBK = -x|a + cj + a + y ( - a + b + cj = ~ ( x + y - ) (1) De H K la doan vuong goc chung cua A B ' va B C t h l : H K l AB' HKIBC HK.AB = , HK.BC = f-(x + y-l)a'+a.b.y + ( y - x ) c ' =0 [(x + y - l ) a ' + y b ' + ( y - x ) c ' - ( x +2y-l)ab = 173 -3x + - y = - — x + 3y = — ^ X = — y =- Thay x, y vao (1) ta duoc: H K = - ( b - c ) = i H K = UK 9V I 9VV / 91 Cflc/i 2: ( Ban doc tu ve hinh ) Goi I , r Ian luot la trung diem ciia B C , B ' C Dat A A ' = h Chpn he true tpa dp Oxyz sac cho O = I, Ox = lA, Oy = IB, Oz = II' Taco: l ( ; ; ) , A Suy ra: A B ' = ^3 ;0;0i , B ^ aVs a , B' f , c y 0;-|;h , B C ' = (0;-a;h) 2 COS(AB',BC') = -^:.h^ AB'.BC BC AB' V^^.V^^ a'+h' = cos ° = 2h a= = a^+h^ ^ h ^ =2a- = ^ h = a72 Vay V,„,,,„,,, = A A ' S _ = a V ^ ^ = ^ 4 Tinh khoang each AB', B C : A B ' = (dvtt) f aV3 //a(-V3;l;2V2), 2 B C = ( ; - a ; a ^ / ) / / b ( ; - l ; ^ / ) ; BB'(0;0;aN/2) a ,b BB' =3.d(AB',BC') = ^ a,b V27 aN/2 aV6 V i du Cho h in h lang t r u tam giac deu A B C A ' B ' C c6 A B = a , goc giira hai mat phSng ( A ' B C ) va ( A B C ) bang ° Goi G la tam tam giac A ' B C Tinh the tich khoi lang t r u da cho va tinh ban kinh mat cau ngoai tiep t u dien GABC theo a ; LOT giai Cdch 1:\ '(A'BC)n(ABC)=BC Goi H la trung diem ciia B C , theo gia thiet ta c6: A ' H I B C A H ± BC 174 (A'BCJTfABC) = X T I X hay = 60° Lai c6: S A B C va A A ' = A H t a n / r H A = — t a n e O " = — Vay the tich khoi lang t r u : a'Vs n « n 3a 3a'V3 (dvtt) Goi J la tam mat cau ngoai tiep t u dien GABC, suy J igiao diem ciia GI voi d u o n g trung true doan GA M la trung diem GA, ta c6: G M G A = GJ.GI => R = GI = g GM.GA GA' 7a GI 2GI 12 Cach 2: Dat A A ' = x,x > Toa cac diem: A(0;0;0),C(0;a;0),A'(0;0;x),C'(0;a;x),B ^aVs Suy A ' B = a 2 ax ax V3 a'Val A'B,A'C , A ' C = (0;a;-x): 2 2 Nen n = ( x ; x V ; a V ) la VTPT cua ( A ' B C ) , k = (0;0;l) l a V T P T c u a (ABC) n.k Theo de bai: - = coseO" =:> iJia = V4x'+3a' 3a a'Vs The tich lang t r u : V , ^ ,,„.c = x.S, ~ x=— 2 3>/3a' as/s _ a a Toa trpng tam G V "T"'2'2 Goi (x; y; z ) la tam mat cau ngoai tiep t u dien G A B C , ta c6 ' ' 2 y2 J 1) l A ' = IC' o ^ x' + y ' + z ' = x' + ( y - a ) ' + z ' l A ' = IB' X +y +z = x~ lA' = IG' x +y" + Tam I a z = aVs X a >'-2; + X z = a y = a z V — 2y z = 12 a , ban kinh R = l A = 7a 12 12 175 V i du Cho lang try A B C D A , B , C , D , c6 day A B C D la hinh chu n h a t , A B = a, H i n h chieu vuong goc ciia diem A , tren mat phang ( A B C D ) triing v o i giao diem A C va B D Goc giua hai mat ph4ng ( A D D , A , ) va ( A B C D ) bang 60° Ti'nh the tich khoi lang try da cho va khoang each t u diem B, den mat phang ( A , B D ) theo a Lai gidi Cdch 1: Gpi I la giao BD=> A,l ( A B C D ) A D ^ I M I A D va diem M cua AC va la trung diem cua va A , M A D , suy la goc gii>a hai mat phang ( A D D , A , ) va ( A B C D ) hay ^ ^ = 60° A,I = I M t a n ^ J V ^ = ^ S^,^„ = AB.CD = a'>/3; 3a^ ^MiCDA.B.C.O, ~ ~ ' Goi B, la d i e m chieu cua B, xuong mat phang ( A B C D ) D o d o d ( B , , A , B D ) chinh la d u o n g cao ve t u B^ cua AOB^B 1 fz = -a.—av3 = CM a^S ^ „ T , T ^ „ aVs ^a-Sl y: B,C // A , D ^ B,C // ( A , B D ) ^ d ( B , , ( A , B D ) ) = d ( c , ( A , B D ) ) D u n g C H B H ( H e B D ) =:> C H ± ( A , B D ) =^ d ( C , ( A , B D ) ) = C H Suyra: d ( B , ; ( A , B D ) ) = C H = ^ £ E £ L = = VCD'+CB- ^ Cdch 2: Goi H la tam cua day A B C D va dat A , H = x Toa dp cac d i e m : A ( ; 0; 0), B(a; 0; 0), D(0; a VS; 0), c(a; a VS; o), H a a ;0 2 suy A A , = a a 2' AA,,AD , A, a a 2 , A D = (0;aN/3;0 r^=(2x;0;a) la VTPT cua ( A , A D ) v a k = (0;0;l) la ax VTPT cua (ABCD) nen theo gia thiet ta c6: 176 n.k Vsa = € " = > 2a = N/4X' + a ' VSa The tich khoi lang t r u : V = x.S ABCD Taco: A,B = a a =^ x = - ^ ^ ^ /- a ' a.av3 = a a -;-x 2' -;-x ,A,D = A,B,A,D = (aV3x;ax;0J Phuongtrinh ( A , B D ) : xVs + y - a V s = V i A,B, = A B = > B , Vay d ( B „ ( A , B D ) ) = 3a a\/3 aVs 2 aVs Ca'c/i 3: Goi O la tarn hinh chii nhat A B C D va M la trung diem A D Ta de dang chung m i n h dugc: A | ( A B C D ) , O M _ L A D v a A,M1AD nen goc giiia hai mat phang ^ A D D ^ A ^ ^ va ( A B C D ) la {A/MyONT)=:\N^ =^ A,O=OM.tan60° = — V s =^P-^AK S u y r a : V^^^.A.B.CD, aVs =— f- 3a^ -.a.av3 = De tinh khoang each t u diem Bi den mat phang (AiBD) ta c6 the chuyen ve ti'nh khoang each giua hai d u o n g thang B,D, va A , B ( v i B,D, //A,BD)) Ta chpn he vecta n h u sau: O D = a; BjA, = b; A , = c v d i = a aVs hayc = De thay X^B^, = 60° nen suy = 60", t u luna y rang a.b = • Taco: B,D, = a va A,B = A , + OB = - a + c Goicacdiem H e B , D , ; K e A , B cho: B,H = xB,D, = 2xa, A , K = y A , B = y | - a + c j H K = HB^ + B / ^ , V A j K = -2xa + b ( - a + cj = - ( x + y)a + b + yc (1) I De H K la doan vuong goc chung cua B|D| va A , B t h i : HK1B,D1 HK ± AlB HK.B,D, = < HK.A,B = 177 3v 2x + y - - + ^ = ^ (2x + y)a- -a.b + yc^ =0 -2(2x + y) + l = -2(2x + y)a' +2a.b = [4x + 2y = l x=— 1^ 7y 2x + - - ^ y =0 2 Thay x,y vao (1) ta dugc: HK = —a + b=>HK = HK 2 =i j — a + b I (A,BD); ; s I T 2 a -a" +a — a = 2 Vi du Cho lang tru ABC.A'B'C c6 dal canh ben bang 2a, day la tarn giac vuong tai A, CO AB = a, AC = aVs Hinh chieu vuong goc cua dinh A' tren mat phang (ABC) la trung diem ciia canh BC Tinh theo a duong cao khoi chop A'.ABC va tinh cosin cua goc giija hai duong thang A A', B'C Lot gidi Cdch 1: Gpi H la trung diem cua BC theo gia thiet A'Hl(ABC) Ta chon cac vecto nhu sau: A ' B ' = a.A'C = b A ' M = c (day la bo ba vecto doi mot vuong goc nen tich v6 huong giua chiing se bSng 0) Tu CO dupe A ' B' = = a, A ' C ' = A'H = = aS, Ta c6:A'A' = A ' H - A H = A T i - ^ ( A B + AC a + bj + c Mat khac A A' = 2a nen -,2 AA " = 4a" - - ( a + b) + c =4a= » - ( a + b)'+c- =4a- c ^ - ( a ^ + b ' ) + c- =4a' cos(AA';B'C')= cos A A ' ; B ' C 4a ^ " • Tinh the tich khoi chop A ' A B C Gpi H la trung diem BQ theo bai ta c6 HK = AM =^ 28 a =^ N K = > / r ^ H t e = 14 , vaa A H = -2 B C = -2V A C ' + A B ' = -2V a ' + a ' =a Do A ' H ' = A ' A - - A H ' = 3a' A ' H = a>/3 • Tinh goc giiia hai d u o n g thang A A ' va Do A A ' / / B B ' , B ' C ' / / B C B'C nen goc giCia hai d u o n g thang A A ' va B ' C giua hai d u o n g thang BB' va BC cung la goc ^ Ap dung dinh l i Pytago cho tam giac A ' B ' H ta c6: H B ' = V A ' B " + A ' H ' = 2a Suyra tam giac B'BH can tai B ' Do ip = '§^BPl la goc s,iua hai duone th3ne A A ' va B ' C :cos(p = =— = * ^ BB' 2.2a Vi du Cho hinh hop d u n g A B C D A ' B ' C ' D ' c6 day la hinh thoi canh a, tam giac ABD la tam giac deu Goi M , N Ian lugt la trung diem ciia cac canh B C , C ' D ' Tinh khoang each t u D den mat phang ( A M N ) biet rang M N J_ B ' D Lai gidi Cdchl: Dai A A ' = x, AB = y, A D = z Ta CO tam giac A B D deu nen: x.y = x.z = 0,y.z = y z cos60" = — Va DB' = D D ' + DC + D A = x + y - z Vi M , N la trung diem ciia BC,C'D' nen M N = M C + CC' + C'N 1 1 ^ hay M N = - A D + CC' + - C ' D ' = - + S - - y ^ 2 2^ Theo bai M N B'D nen M N D B ' - n a' Do (x + y - z ) - z + x — y {2 2\ = < = > x " + - 2 2 a — a 2 a' „ V2 +-.— = o x = —a 2 179 Ta c6: d ( D , ( A M N ) ) = ^ ^ ^ D'e thay S , , „ = "'AMN (dvdt) 2V83 Goi H la trung diem cua D C thi N H (ABCD), N H = ^ a nen V ^D.AMN = V *N.AMD ^ - N H S p^'^'^'^AMD 7^ 24 Ke H K ± A M ta c6 N K ± A M Theo d j n h l i cosin ta c6: A M ' =BA'+BC'-2BA.BC.cosl20° Ta C O S^j,„ - S^^p Nen H K = ^ ^ ^ = AM /v = - a ' => A M = — a (SADH + ^CHM + ^ABM ) ~ Q ^ 28 3S 16 a^ >NK = V N H ' H K ' = ^ a 14 2v'6 Suy S^^^ = - N K A M = ^ a " - d e do-d(©^,(AMN)) = ^ a V83 2N/6 Vay khoang each l u diem D den mat phang ( A M N ) la - = a V83 Cdch 2: Goi O; O ' Ian lugt la tam ciia hinh thoi ABCD va A ' B ' C ' D ' Ta c6: BD = a; A C = A = ^ = aS Dat A A ' = h Chon he true tga Oxyz v o i O ( , , ) ; Ox = OA; O y ^ O D ; Oz = 0 ' Ta c6: A US ;0;0 B' ; - - ; h ; C' I , •MN ; aVs ;C -;0;0 ;D' I ;D a 0;-;0 , ^ ;B ;M 0; ;0 aVa ^ h-=Ooh = — = (V2;-3>/6;3V3) U,,U2 ; N B'D(0;a;-h) Do M N B'D nen M N B ' D = o '3^3 a ^ A M — - - a ; - - - ; //u;(3V3;l;0) 4 MN 180 Mat phSng ( A M N ) qua A ( 4~2 X a/3 ;0;0 VTPT CO n(AMN) CO p h u a n g trinh: aVs V2x„-3V6yp+3N/3z„ ^ Suy ra: d ( D , ( A M N ) ) = 2V6 83 2) +(-3V6)"+(3V3)' a V i d u Cho hinh hop d i i n g ABCD.A'B'C'D' c6 day la hinh vuong, tarn giac A ' A C vuong can, A ' C = a Tinh the tich cua khoi t u dien ABB'C'va khoang each tie diem A den mat phSng (BCD') theo a Lai gidi Cdch 1: (Ban doc tu ve hinh) Do AAA C vuong can tai A nen ta tinh duoc: A A ' = A C = va canh ciia hinh vuong AB = - Theo de bai A B I B B ' v a A B I B C nen de dang suy duoc: A B l ( B B C C ) A u c a a a^/2 a'Vz buy ra: V.„„.^ =-.AB.b.„„,,,, = - - — - = y ABBc ^""^ 2 2 48 De tinh khoang each t u A den mat phang (BCD') ta c6 the tinh khoang each giiia hai duong duong A D va CD' ( v i ( A D // (BCD)) Ta chpn he vecta n h u sau: B A = a ; A D = b ; A A ' = c voi a ayfl hay c' = —; c = Taco: C D ' = C C + C ' D ' = a + c Goi cac diem H e A D va K G C D ' cho: A H = x A D = xb; CK = yCD^' = y (a + c ) H K = H A + AC + A K = H A + BC~BA + A K = - x b + b - a + y|a + cj = ( y - l ) a + ( l - x ) b + yc (1) De H K la doan vuong goe chung cua A D va CD' thi: l~x = y - l + 2y = fHK AC HK.AC = HKICD' HK.CD' = x=l 1 => H K = 2a + c (the x, y vao t u (1)) y =3 3l I 181 HK = HK -2a+ c V = ^ V a y d[A;(BCD') = HK = J 76 Cdch 2: (Ban dgc tu ve hinh) Ta C O A A ' A C v u o n g can nen A A ' = A C = V.HHc ABBc Lai c6 A C = A B V => A B = - =-AB.S,,,,, = l A B l B B ' B ' C ' = l ^ ^ ^ = ^ ( d v t t ) ABBt 2 48 ^ ^ Taco: ( B C D ' ) = ( B C D ' A ' ) Trong A A B A ' , d u n g A H A ' B D o B C ( A B B ' A ' ) ^ B C A H => A H ( B C D ' A ' ) ^ d(A,(BCD')) = d(A,(BCD'A')) =A H Xet A A B A ' ( A = 90"), A H A ' B : AH- •+- AH = AB' A A " J 76 BAI TAP lif LUYEN Bai tap 1: Cho h i n h chop S.ABCD c6 SA = a72 va tat ca cac canh lai c6 d p dai bang a Chung minh duong thang BD vuong goc voi mat phSng (SAC) va ti'nh the ti'ch khoi chop S.ABCD theo a Cho hinh chop tam giac S.ABC c6 AB = 5a, BC = 6a, CA = 7a(a > O) Cac mat ben SAB,SBC,SCA tao voi day mot goc 60" Tinh the tich ciia tchoi chop S.ABC theo a Trong khong gian, cho tam giac vuong can A B C c6 canh huyen AB = 2a(a > ) Tren d u o n g thang d d i qua A va vuong goc voi mat phang ( A B C ) lay diem S cho mat phSng ( S B C ) tao voi mat phSng ( A B C ) mot goc bang ° Tinh dien tich mat cau ngoai tiep t u dien S A B C Cho hinh chop S.ABCD c6 day ABCD la hinh vuong canh a H i n h chieu vuong goc ciia S len mat phang (ABCD) triing v o i tam tam giac A B D Canh SD tao voi day (ABCD) mot goc bSng " T i n h the tich khoi chop S.ABCD va khoang each t u A toi mat phang (SBC) theo a Cho hinh chop S A B C D c6 day la hinh chCr nhat A B C D c6 AB = 2a, A D = 4a, SA ( A B C D ) va • ( S C ^ ( A B C D ) ) = 30° T i n h the tich cac h i n h chop S A B C D ; S A H M N va khoang each giiia M N va SB Cho h i n h chop S.ABCD c6 day ABCD la hinh vuong canh a, SB vuong goc voi day, SB = a Gpi P la diem thoa man DP = ^ BS a) T i n h the tich ciia khoi t u dien ACSP 182 b) Chung m i n h mat phang ( A C S ) vuong goc v o i mat phang ( A C P ) c) Ti'nh goc va khoang each giua BC va SP Cho hinh chop S.ABCD c6 day la hinh vuong canh a, SA v u o n g goc v o l mat phang day, SA = a G o i M , N Ian lugt la trung diem cua cac canh SB va SD; I la giao diem cua SC va mat phang ( A M N ) a) Hay neu each xae d i n h diem I b) Chung m i n h rang SC vuong goe A I c) Tinh the tich khoi chop S M A N I Cho h i n h chop S A B C c6 day la tam giac v u o n g can tai C, canh h u y e n bang 3a, G la tam tam giac A B C , SG ± ( A B C ) , SB = • Tinh the tich khoi chop S A B C va khoang each t u B den mat phang ( S A C ) theo a Cho hinh chop S.ABC, day ABC la tam giac vuong tai B c6 AB = a;BC = 2a; SA vuong goc v d i mat phang (ABC) Goi M la trung diem cua A C , biet goc giiia mat phang (SBC) va (ABC) bang ° Tinh the tich khoi chop S.ABC va khoang each t u C den mat phing (SBM) theo a 10 Tren d u o n g thang vuong goc tai A v d i mat phang cua h i n h vuong A B C D canh a, lay diem S vdi SA = 2a Goi B ' , D ' la hinh chieu vuong goc cua A len SB va SD Mat phang ( A B ' D ' ) eat SC tai C Tinh the tich khoi da dien A B C D D ' C ' B ' Bai tap 2: Cho hinh chop t u giac S A B C D cd hai mat ( S A C ) va ( S B D ) cung vuong goc v d i day, day A B C D la hinh ehu" nhat cd AB = a,BC = ajs , diem I thuoc doan thang SC cho SI = 2CI va thoa man A I _L SC Tinh the tich cua khoi chdp S A B C D theo a Cho hinh chdp S.ABC cd ABC va SBC la cac tam giac deu canh a, SA = Tinh theo a khoang each iii d i n h B den mat phang ( S A C ) Cho hinh chdp S A B C D , cd day A B C D la hinh thang can, day Idn A D = 2a, AB = B C = a, SB = 2a , hinh chieu vuong gdc ciia S tren mat phSng ( A B C D ) triing v d i trung diem O cua A D Tren eae canh SC, SD lay diem M , N cho S M = M C , SN = O N Mat phSng ( a ) qua M N va song song v d i BC cat SA, SB Ian lugt tai P, Q Tinh the tich khoi chdp S.MNPQ theo a Cho hinh chdp S A B C D cd S A = aVs , h i giac A B C D la hinh thang can vdi day Idn la AD, AB = BC = C D = a, • ^ A b = 60'' Hinh chieu vuong gdc eiia S tren mat ph3ng ( A B C D ) thupe doan thang A D , mat ben ( S A B ) tao vdi mat phang ( A B C D ) mot gdc ° Tinh theo a the tich khoi chdp S A B C D Cho hinh chdp S A B C D cd day A B C D la hinh thoi tam O; tam giac S B D deu canh 2a, tam giac S A C vuong tai S cd S C = aVs; gde giCra ( S B D ) va mat day la 60° Tinh theo a the tich khoi chdp S A B C D va khoang each giiJa d u d n g thang A C va d u d n g thang SB 183 Cho hinh chop S.ABCD c6 S C ( A B C D ) , day ABCD la hinh thoi c6 canh bang aVs va :^B^ = 120° Biet rang goc giua hai mat phang (SAB) va (ABCD) bSng 45° Tinh theo a the tich khoi chop S.ABCD va khoang each giira hai dudng thang SA, BD Cho h\nh chop S.ABCD c6 day ABCD la hinh thoi tam O Biet A C = 2a; BD = laS H i n h chieu d i n h S len mat phSng (ABCD) la trung diem H cua OB Goc giira hai mat phang (SCD) va (ABCD) b3ng 60^ Tinh the tich khoi chop S.ABCD va khoang each giiia hai duong thSng SB va A D Cho hinh chop S.ABCD c6 day ABCD la nua luc giac deu npi tiep duong tron duong k in h A D = 2a, SA vuong goc v o i day Goi M la trung diem canh SB, biet khoang each til B den mat phang (SCD) bang • Tinh the tich khoi chop S.ABCD va cosin goc giua hai d u o n g thSng BC va D M Cho hinh chop deu S.ABCD c6 day ABCD la hinh vuong tam O, AB = a, OS goc giiia hai mat phang (SAB) va (SCD) bang 60° Goi M la trung diem cua canh SC Mat phang (ABM) cat SD tai diem N Tinh theo a the tich cua khoi chop S.ABMN va chung m i n h rSng hai mat phang (ABM) va (SCD) vuong goc voi 10 Cho hinh chop S.ABCD c6 day ABCD la hinh binh hanh AB = ; BC = 2a va tam giac A B D can tai B Goi G la trpng tam ciia tam giac A B D ; M , N Ian lupt la trung diem cua SB, SC va SG = SB = SC Biet goc tao boi hai mat phSng (SBC) va ( A B C D ) la 60° Tinh theo a the tich khoi chop G M N D va khoang each giua hai duong thSng D M , SN Bai tap 3: Cho hinh chop S.ABC c6 day ABC la tam giac vuong tai A, AB = a, A C = aVs Gpi I la trung diem canh BC, hinh chieu vuong goc cua S xuong mat ph3ng (ABC) la trung diem H cua A l Goc giua hai mat phSng (SAB) va (ABC) bang 6O0 Tinh the tich S.ABC va khoang each hai d u o n g thang SC va A I Cho hinh chop S.ABC c6 day ABC la tam giac deu canh 2a\/3, mat ben SAB la tam giac can voi Xs6 = 120" va nam mat phang vuong goc v o i day Gpi M la trung diem ciia SC va N la trung diem cua MC Tinh the tich khoi chop A M N B va khoang each giua hai d u o n g th5ng A M , B N Cho hinh chop S.ABC c6 day la tam giac ABC vuong tai C, AB = 5cm, BC = 4cm Canh ben SA vuong goc vai day va goc giira canh ben SC voi mat day (ABC) b3ng 60° Gpi D la trung diem ciia canh AB Tinh the tich khoi chop S.ABC va khoang each giiia hai duong thang SD va BC Cho hinh chop S.ABCD c6 day ABCD la hinh thang vuong tai A va D, AB = A D = 2a, C D = a Canh ben SA vuong goc voi mat phang day, mat phang (SBC) tao voi mat ph5ng day goc 60° Gpi M la trung diem SB, mat ph4ng ( A D M ) cat canh sc tai N Tinh the tich khoi chop S A B C D va khoang each giiia hai duong thSng B N va C D theo a 184 Cho hinh chop S.ABCD c6 day ABCD la h i n h thoi, AB = 2a; "SAb = ° H i n h chieu vuong goc ciia d i n h S len mat phang (ABCD) la tarn H cua tarn giac A B D Biet tarn giac SAC vuong tai d i n h S, tinh the tich khoi chop S.ABCD theo a va tinh goc giua hai mat phang (SAC) va (SBC) Cho hinh chop S.ABCD c6 day ABCD la hinh vuong canh a, tam giac SAB deu va nam mat phSng vuong goc v d i day Goi M , N , P, K Ian l u g t la trung diem cua BC, CD, SD, SB Tinh the tich cua khoi chop S.ABMN va khoang each giua hai d u o n g thang MK va AP theo a Cho hinh chop S.ABCD c6 day la hinh vuong canh a, SB = SD Biet rSng dien tich cac tam giac SBD va SAD Ian lugt la 3a' , a-^Iv7 va — - — , dong thoi cac mat ben la n h i i n g tam giac nhon Tinh the tich khoi chop S.ABCD va khoang each giua hai d u o n g thang BC va SD Cho hinh chop S.ABCD c6 day ABCD la h i n h chu' nhat canh AB = a; BC = aV2 Canh ben SA vuong goc v o i day Goi M , N Ian lugt la trung diem ciia SD va A D Mat phang (P) chua B M c§t mat ph3ng (SAC) theo mot d u o n g thang vuong goc v o i B M Gia sii' BN cSt A C tai I , goi ] la trung diem ciia IC Biet khoang each t u d i n h S den mat phang (P) bSng ^ ' ^ ^ Tinh the tich khoi chop BMDJ va khoang each giua hai d u o n g thang D M va BJ theo a Bai tap 4: Cho lang try dung ABC.A B C c6 day ABC la tam giac vuong voi AB = BC = a , canh ben A A = a , M la diem cho A M = ~ AA Tinh the tich cua khoi t u dien M A BC Cho hinh lang tru d i i n g ABC.A^B^C^ c6 day la tam giac deu Mat phang ( A ^ B C ) tao voi day mot goc 30 va tam giac A j B C c6 dien tich hang 18 Tinh the tich khoi lang try A B C A ^ B ^ C j Trong khong gian cho lang try d u n g A B C A ^ B j C j c6 AB = a, A C = 2a, AA^ = 2aV5 va "SAt: = 120° Goi M la trung diem cua canh C C , Chung m i n h M B M A j va tinh khoang each t u A toi mat phang ( A ^ B M ) Cho hinh lang try ABCA^B^Cj c6 M la trung diem canh AB, BC = 2a, goc ACB bSng 90° va goc ABC bSng 60°, canh ben CC^ tao vol mat phSng (ABC) mot goc 45°, hinh chieu vuong goc ciia C, len mat phang (ABC) la trung diem ciia C M Tinh the tich khoi lang try da cho va goc tao boi hai mat phang (ABC) va ( A C C j A ^ ) Cho lang try d u n g A B C A ' B ' C c6 day la tam giac deu Goi M la trung diem cua canh BB' Biet hai d u o n g thSng A ' B , C M vuong goc voi va each mot khoang hang ^^jj^- Tinh theo a the tich khoi lang try A B C A ' B ' C 185 ... x^+2tx^+3t^x2 =17 + m 11 + 2t + t '' ( l + 2t + 3t2) 11 + 2t + t ^ = 17 + m + 2t + t^ ( m - ) t ^ + ( m + 6)t + m + 40 = CO 3x^+2tx2+t^x^ =11 11 x2= Ta m + 17 3y^ = m + 17 11 + 2t + t '' > 0,Vt e... max (t,\i ^ 405 216 0 ,^ f t ) = f 2 013 1 = 2 013 t6(0;2003| ^ '' ^ '' „ „ , = 2 013 2 013 , d a u = x a y k h i : x = y = z = 2 013 ^ ^ ^ 405 216 0 2 013 Vay m a x A = , x = y = z = 2 013 Ta c6: P = (a... - x = x ^ - x + 40 (>/^-AAr72)(^l +Vx^ + X + 10 = V7-X |5-x V25-x^ - V l O - x ^ = 11 ^2^ + 4^ =1 10 ^/iWT + ^1- N/^ = 12 x^ - X - lOOoVl + SOOOx = 10 00 39 Bai tap 2: Giai bat p h u a n g trinh