BI TP THC HNH S DNG PHN MM HèNH HC NG TRONG MT PHNG Kim tra gi thuyt, tỡm hiu v khỏm phỏ bi toỏn: 1) Trờn on thng AB ta ly mt im C nm gia A v B Dng cỏc tam giỏc u ACE v BCF cho E, F nm cựng phớa i vi ng thng AB a) So sỏnh di hai on thng AF = BE b) Gi M v N ln lt l trung im cỏc on thng AF v BE Khi ú, CMN cú c im gỡ? 2) Trờn ba cnh ca ABC cho trc, ta dng phớa ngoi ba tam giỏc u ABC, BCA, CAB a) Cú nhn xột gỡ v di cỏc on thng AA, BB, CC ? b) Gi O1, O2, O3 l tõm ba tam giỏc u trờn Khi ú O1O2O3 cú c im gỡ? c) Xột bi toỏn nu dng phớa ngoi ABC cỏc tam giỏc cõn 3) Cho hỡnh bỡnh hnh ABCD Cỏc ng phõn giỏc ca cỏc gúc ca hỡnh bỡnh hnh ct to thnh mt t giỏc a) D oỏn v hỡnh dng ca t giỏc to thnh trờn b) Xột bi toỏn nu ABCD l hỡnh ch nht 4) Cho ng trũn (O) cú ng kớnh AB Gi C l im i xng vi A qua B v PQ l ng kớnh thay i ca (O) khỏc ng kớnh AB ng thng CQ ct PA v PB ln lt ti M, N a) Kim tra cỏc ng thc sau: QC = QM, NC = NQ b) Minh qu tớch cỏc im M, N ng kớnh PQ thay i 5) Cho M l mt im nm trờn ng trũn ngoi tip ABC Gi M1, M2, M3 ln lt l cỏc im i xng ca M qua cỏc cnh BC, CA, AB v gi H l trc tõm ca ABC Hóy nhn xột v v trớ ca cỏc im M1, M2, M3, H 6) Cho t giỏc li ABCD Dng phớa ngoi t giỏc hỡnh vuụng ABB'A', BCC'B", CDC"D', DAD"A" Gi tõm ca hỡnh vuụng theo th t ú l E, F, G, H Hóy nhn xột v hỡnh dng ca t giỏc to bi trung im ca cỏc ng chộo ca hai t giỏc ABCD, EFGH 7) Cho ng trũn (O) v dõy cung AB Gi C l trung im ca AB, qua C k hai cỏt tuyn tu ý l HI v KJ Gi M, N l giao ca AB vi IJ, KH Hóy kim tra ng thc CM = CN Kt qu ny ca bi toỏn cũn ỳng khụng nu ta thay ng trũn bng ng elớp? 8) Cho ABC ni tip (O) Gi H, G, O, O9 ln lt l trc tõm, trng tõm, tõm ng trũn ngoi tip v tõm ng trũn le ca ABC Nhn xột gỡ v v trớ ca bn im trờn? 9) Cho tam giỏc ABC ni tip ng trũn (O), trc tõm H Gi A, B, C ln lt l cỏc im i xng ca H qua BC, CA, AB Gi O1, O2, O3 l tõm ba ng trũn ngoi tip cỏc tam giỏc HBC, HCA, HAB Hóy kim tra gi thuyt: O1O2O3 = ABC 10 Trờn ng trũn (O) ly hai im B, C c nh v im A thay i Gi H l trc tõm ca ABC v H l im cho HBHC l hỡnh bỡnh hnh Chng minh rng H nm trờn ng trũn (O) Minh qu tớch, tỡm hiu bi toỏn qu tớch: Cho hai ng trũn (O) v (O) tip xỳc ti A Gi AB l ng kớnh ca ng trũn (O) v AC l ng kớnh ca ng trũn (O) Mt ng thng thay i i qua A ct hai ng trũn (O) v (O) ln lt ti M v N Tỡm qu tớch giao im ca BN v CM Cho ng trũn (O) cú AB l mt ng kớnh c nh v M l mt im chy trờn ng trũn (O) Trờn tia AM ly im I cho MI = MB Tỡm qu tớch im I Cho hai im A, B v ng trũn tõm O khụng cú im chung vi ng thng AB Qua mi im M chy trờn ng trũn (O), dng hỡnh bỡnh hnh MABN Chng minh rng, im N thuc mt ng trũn xỏc nh Cho ABC ni tip ng trũn (O) v D l mt im chuyn ng trờn cung BC khụng cha nh A H CH vuụng gúc vi AD Tỡm qu tớch ca im H Cho na ng trũn ng kớnh AB v M l mt im chuyn ng trờn na ng trũn ú V tam giỏc vuụng cõn MBC (BM = BC) ngoi AMB Tỡm qu tớch im C Cho AC l mt dõy cung bt kỡ ca na ng trũn tõm O ng kớnh AB = 2R K tip tuyn Ax vi na ng trũn y Tia phõn giỏc ca gúc CAx ct ct tia BC D Tỡm qu tớch ca D Cho hỡnh vuụng ABCD v mt im M di ng trờn ng chộo BD Dng ME vuụng gúc vi AB v MF vuụng gúc vi AD Tỡm qu tớch giao im P ca CF v DE, giao im N ca CE v BF Cho na ng trũn ng kớnh AB, C thuc cung AB; dng hỡnh vuụng CBEF phớa ngoi tam giỏc ABC Tỡm qu tớch im E Cho ng trũn (O) vi ng kớnh AB c nh, mt ng kớnh MN thay i Cỏc ng thng AM v AN ct cỏc tip tuyn ti B ln lt ti P v Q Tỡm qu tớch trc tõm ca cỏc tam giỏc MPQ v NPQ 10 Tam giỏc ABC cú hai nh B, C c nh cũn nh A chy trờn mt ng trũn (O; R) c nh khụng cú im chung vi ng thng BC Minh qu tớch trng tõm G ca tam giỏc ABC Minh im c nh ca h ng cong: 1) (Cm) : y = x3 mx2 (2m2 7m + 7)x + 2(m 1)(2m 3) 2) (Cm) : y = 3) 4) 5) (Cm) : y = x 2m x + 3mx + 2m 3m + 2 (m + 1)x + m + x +m + 2x2 + (1- m)x + (1+ m) (Cm) : y = x- m mx2 + 2(m + 1)x + 3m2 - m (Cm) : y = x +1 (Cm) : y = (m - 1)x + m + x +m + (Cm) : y = 2x2 - (6- m)x + mx + 6) 7) (Cm) : y = x3 (m + 1)x2 (2m2 3m + 2)x + 2m(m 1) 8) Minh s giao im ca th hm s (C) vi ng thng (Cm): y= 1) (C) y= 2) (C) x x +1 x+3 x +1 3) (C) 4) (C) 5) (C) 6) (C) 7) (C) 8) (C) v (Cm) v (Cm) y = x3 + 3x y= v (Cm) v (Cm) y = x3 + 3x2 y = x2 2x y = x2 + x y=m v (Cm) v (Cm) y = x4 + 2x2 + , vi m l tham s y = 2x + m y = x + 3x + 9) Trc ox v y = mx m , vi m l tham s , vi m l tham s y=m y=m v (Cm) v (Cm) , vi m l tham s , vi m l tham s , vi m l tham s y=m , vi m l tham s y = 3x + m , vi m l tham s (Cm) : y = 2x2 + 2mx + m , vi m l tham s y= 10) (C) 11) (C) x2 4x + x v (Cm) y = x3 3x2 + v (Cm) y = x + m , vi m l tham s y = m(x 1) , vi m l tham s (Cm) : y = x2 + (2m + 1)x + m2 1, m Minh h ng cong ng thng c nh luụn tip xỳc vi mt BI TP THC HNH S DNG PHN MM HèNH HC NG TRONG KHễNG GIAN Minh qu tớch, tỡm hiu bi toỏn qu tớch Cho hỡnh chúp ỏy l hỡnh vuụng ABCD cnh a, SA = a v vuụng gúc vi mt phng (ABCD) Gi M l im bt kỡ trờn on BC, gi K l hỡnh chiu ca S lờn DM Tỡm qu tớch im K Cho tam giỏc u SAB v hỡnh vuụng ABCD cnh a nm hai mt phng vuụng gúc vi Gi H, K ln lt l trung im ca AB, CD Gi M l im di ng trờn on SA Tỡm qu tớch im P l hỡnh chiu ca S trờn mt phng (CDM) Cho hỡnh chúp S.ABCD cú ỏy ABCD l hỡnh vuụng v SA vuụng gúc vi ỏy ti A Trờn AD ly im M bt kỡ Gi I l trung im ca SC, H l hỡnh chiu ca I lờn CM Tỡm qu tớch ca im H Cho hỡnh lng tr ng ABC.ABC ỏy l tam giỏc cõn vi AB = AC = a, BC = b v AA = h Gi P l im bt kỡ trờn on AA, E l hỡnh chiu ca C lờn BP Tỡm qu tớch ca E Dng thit din, tỡm hiu hỡnh dng thit din: Cho hỡnh lp phng ABCD.ABCD, im M di ng trờn tia AC; im N, K ln lt nm trờn on thng AB, AD Xỏc nh thit din to bi mt phng i qua im M, N, K v cho bit hỡnh dng ca thit din ú ng vi nhng v trớ ca M Cho hỡnh chúp S.ABCD vi ỏy ABCD l hỡnh thang ABCD cú AD song song vi BC v AD = 2BC Gi E l trung im AD v O l giao im ca AC v BE I l im di ng trờn cnh AC khỏc vi A v C Qua I, ta v mt phng () song song vi (SBE) Tỡm thit din to bi () v hỡnh chúp S.ABCD Cho hỡnh hp ABCDABCD Gi I l mt im thuc on AB Xỏc nh thit din to bi hỡnh hp v mt phng () qua I, song song vi BD v AC 4 Cho hỡnh chúp S.ABCD cú ỏy l hỡnh vuụng cnh a, SA vuụng gúc vi ỏy, SA cú di bng a v M thuc AC Xỏc nh thit din to bi hỡnh chúp S.ABCD vi mt phng () l mt phng qua M v song song vi SA, BD Cho hỡnh lp phng ABCDEFGH, im M di ng trờn on thng EG Xỏc nh thit din to bi mt phng i qua im M v vuụng gúc vi AG BI TP THC HNH S DNG PHN MM MAPLE S dng phn mm v dóy im minh hỡnh nh gii hn ca cỏc dóy s: 1) un = n+3 un = sin n + n sin n un = ( 1) + n+3 n , 2) , 3) un = sin n + n n un = , 4) sin n n +1 , 5) n Cú th s dng cỏc cõu lnh sau: [> with(plots): [> pointplot([seq([(n^2+1)/(2*n^2-2),0],n=2 100)],symbol=box,color=black); [> pointplot([seq([n,(-1)^n/n],n=1 100)], symbol=cross,color=red); [> pointplot([seq([(3*n^2+4*n-7)/(n^2),0],n=1 200)], symbol=circle,color=red); S dng phn mm i s gii cỏc phng trỡnh sau: 2 2x x 22+xx = 1) ; pt:=2^(x^2-x)-2^(2+x-x^2)=3; solve(pt,x); x + + x +1 x +1 = 2) pt:=2*sqrt(x+2+2*sqrt(x+1))-sqrt(x+1)=4; solve(pt,x); ) ( ( x + ) +1 x 2 =0 3) pt:=(sqrt(2)-1)^x+(sqrt(2)+1)^x-2*sqrt(2)=0; solve(pt,x); 2x - + x2 - 3x + = 4) pt:=sqrt(2*x-1)+x^2-3*x+1=0; solve(pt,x); 23 3x - + 6- 5x - = 5) pt:=2*(3*x-2)^(1/3)+3*sqrt(6-5*x)-8=0; solve(pt,x); 3x + - - x + 3x2 - 14x - = 6) pt:=sqrt(3*x+1)-sqrt(6-x)+3*x^2-14*x-8=0; solve(pt,x); S dng phn mm i s gii cỏc bt phng trỡnh sau: (x ) 2x2 - 3x - 5x - - x - > 2x - - 3x 1) bpt:=(x^2-3*x)*sqrt(2*x^2-3*x-2)>=0; solve(bpt,{x}); 2) bpt:=sqrt(5*x-1)-sqrt(x-1)>sqrt(2*x-4); solve(bpt,{x}); ( ) x2 - 16 x- + x- 3> 7- x x- 3) bpt:=sqrt(2*(x^2-16))/sqrt(x-3)+sqrt(x-3)>(7-x)/sqrt(x-3); solve(bpt,{x}); S dng phn mm i s gii cỏc h phng trỡnh sau: ỡù ùù x - = y - ùớ x y ùù ùùợ 2y = x + 1) pt1:=x-1/x=y-1/y; pt2:=2*y=x^3+1; solve( {pt1,pt2}, {x,y} ); ỡù x ( x + y + 1) - = ùù ùù ( x + y) - + = ùùợ x2 2) pt1:=x*(x+y+1)-3=0; pt2:=(x+y)^2-5/x^2+1=0; solve( {pt1,pt2}, {x,y} ); ỡù x - y = x - y ùù ùù x + y = x + y + ùợ 3) pt1:=(x-y)^(1/3)=sqrt(x-y); pt2:=x+y=sqrt(x+y+2); solve( {pt1,pt2}, {x,y} ); ỡù x + y - xy = ùù ùù x + + y + = ùợ 4) pt1:=x+y-sqrt(x*y)=3; pt2:=sqrt(x+1)+sqrt(y+1)=4; solve( {pt1,pt2}, {x,y} ); ỡù xy + x + y = x2 - 2y2 ù ùù x 2y - y x - = 2x - 2y ùợ 5) pt1:=x*y+x+y=x^2-2*y^2; pt2:=x*sqrt(2*y)-y*sqrt(x-1)=2*x-2*y; solve( {pt1,pt2}, {x,y} ); ỡù xy + x + = 7y ù 2 ùù x y + xy + = 13y2 ợ 6) pt1:=x*y+x+1=7*y; pt2:=x^2*y^2+x*y+1=13*y^2; solve( {pt1,pt2}, {x,y} ); ỡù x4 + 2x3y + x2y2 = 2x + ù ùù x + 2xy = 6x + ùợ 7) pt1:=x^4+2*x^3*y+x^2*y^2=2*x+9; pt2:=x^2+2*x*y=6*x+6; solve( {pt1,pt2}, {x,y} ); ỡù ùù x + y + x3y + xy2 + xy = - ùớ ùù ùù x + y + xy ( 1+ 2x) = ùợ 8) pt1:=x^2+y+x^3*y+x*y^2+x*y=-5/4; pt2:=x^4+y^2+x*y*(1+2*x)=-5/4; solve( {pt1,pt2}, {x,y} ); ỡù ùù 3y = y + ùù x2 ùù x +2 ùù 3x = y2 ùợ 9) pt1:=3*y=(y^2+2)/x^2; pt2:=3*x=(x^2+2)/y^2; solve( {pt1,pt2}, {x,y} ); S dng phn mm i s tớnh cỏc tớch phõn sau: ũ ln( x ) - x dx 1) int( ln(x^2-x), x=2 ); e ũ 2) 1+ 3ln x ln x dx x int( (sqrt(1+3*lnx)*lnx)/x, x=1 e ); x ũ 1+ x- 1 3) dx int( x/(1+sqrt(x-1)), x=1 ); p sin2x + sin x ũ 4) 1+ 3cosx dx int( (sin2x+sinx)/(sqrt(1+3*cosx)), x=0 Pi/2); dx ũx x2 + 5) int( 1/(x*sqrt(x^2+4)), x=sqrt5 2*sqrt3 ); p sin2x cosx ũ 1+ cosx dx 6) int( (sin2x*cosx)/(1+cosx), x=0 pi/2 ); p ũ( e sin x 7) ) + cosx cosxdx int( (e^sinx+cosx)*cosx, x=0 pi/2 ); p ũ sin2x 2 cos x + 4sin x dx 8) int( sin2x/sqrt(cosx^2+4*sinx^2), x=0 pi/2 ); ũ( x - 2) e 2x dx 9) int( (x-2)*e^2*x, x=0 ); e ũx ln2 xdx 10) int( x^3*lnx^2, x=0 e ); ũ 11) lnx dx x3 int( lnx/x^3, x=1 ); p ũ( cos x - 1) cos xdx 12) int( (cosx^3-1)*cosx^2, x=0 pi/2 ); ũ + ln x ( x + 1) dx 13) int( (3+lnx)/(x+1)^2, x=1 ); dx x - ũe 14) int( 1/(e^x-1), x=1 ); x2 + ex + 2x2ex ũ 1+ 2ex dx 15) int( (x^2+e^x+2*x^2*e^x)/(1+2*e^x), x=0 ); e ũ ln x x ( + ln x) dx 16) int( lnx/(x*(2+lnx)^2), x=1 e ); ổ 3ử ữln xdx ỗ x ỗ ữ ũ ỗố xứữ ữ e 17) int( (2*x-3/x)*lnx, x=1 e ); x2 + ex + 2x2ex ũ 1+ 2ex dx 18) int( (x^2+e^x+2*x^2*e^x)/(1+2*e^x), x=0 ); S dng phn mm i s tỡm giỏ tr ln nht, giỏ tr nh nht ca hm s sau: y = x + - x2 1) minimize(x+sqrt(4-x^2), x); maximize(x+sqrt(4-x^2), x); y = - x2 + 4x + 21 - - x2 + 3x + 10 2) minimize(sqrt(-x^2+4*x+21)-sqrt(-x^2+3*x+10), {x}); maximize(sqrt(-x^2+4*x+21)-sqrt(-x^2+3*x+10), {x}); x +1 y= x2 + 1; 3) trờn on minimize((x+1)/sqrt(x^2+1),{x}, {x=-1 2}); maximize((x+1)/sqrt(x^2+1),{x}, {x=-1 2}); y = 5- 4x 4) trờn on [-1; 1] minimize(sqrt(5-4*x),{x}, {x=-1 1}); maximize(sqrt(5-4*x),{x}, {x=-1 1}); y = x4 - 3x2 + 5) trờn on [0; 3] minimize(x^4-3*x^2+2,{x}, {x=0 3}); maximize(x^4-3*x^2+2,{x}, {x=0 3}); y= 2- x 1- x 6) trờn on [2; 4] minimize((2-x)/(1-x),{x}, {x=2 4}); maximize((2-x)/(1-x),{x}, {x=2 4}); S dng phn mm i s tớnh tng v tớch vụ hn sau: k k =1 1) , F=sum('1/k^2', 'k'=1 infinity); F=product( 1/k^2, k=1 infinity ); Ơ k=0 k ! 2) F=sum('1/k!', 'k'=0 infinity); F=product( 1/k!, k=0 infinity ); k=1 ữ 4k 3) Product( 1-1/(4*k^2), k=1 infinity ) = product( 1-1/(4*k^2), k=1 infinity ); S dng phn mm i s tớnh o hm cp 1, o hm cp ca cỏc hm s sau: ( ) y = ln x + x2 + 1) dhb1:=diff(ln(x+sqrt(x^2+1)),x); dhb2:=diff(ln(x+sqrt(x^2+1)),x$2); 2) ( y = tan ln( x) ) dhb1:=diff(tan(ln(x)),x); dhb2:=diff(tan(ln(x)),x$2); y= 3) 23x 32x dhb1:=diff(2^(3*x)/3^(2*x),x); dhb2:=diff(2^(3*x)/3^(2*x),x$2); 4) y = xx dhb1:=diff(x^(x^2),x); dhb2:=diff(x^(x^2),x$2); x 5) y = x + xx + xx dhb1:=diff(x+x^x+x^(x^x),x); dhb2:=diff(x+x^x+x^(x^x),x$2); y=x x 6) dhb1:=diff(x*abs(x),x); dhb2:=diff(x*abs(x),x$2); S dng phn mm i s tớnh gii hn ca cỏc hm s sau: a) 3.2n - (1)n y= 2n limit((3.2^n-1^n)/2^n, x=infinity); 3n+2 b) ổ n + 1ữ ữ y =ỗ ỗ ữ ỗ ố n ữ ứ limit(((n+1)/n)^(3*n+2), x=infinity); y= c) sin( 1+ 3n) + n2 limit(sqrt(9/4+(sin(1+3*n))/n^2), x=infinity); 10 S dng phn mm Maple vit chng trỡnh : 1) Kim tra gi thuyt toỏn hc: a) Nu kớ hiu pk l s nguyờn t th k (k t s tr i) thỡ mt s nguyờn t vi mi n n Fn = 2 + An = p1 p2 pn b) Vi mi s nguyờn dng n, ta luụn cú l mt s nguyờn t 2) Kim tra v trớ tng i ca hai ng thng mt phng 3) Gii h phng trỡnh bc nht hai n 11 S dng phn mm Maple vit chng trỡnh kho sỏt hm s a) y = ax3 + bx2 + cx + d, vi a, b, c, d nhp t bn phớm restart: with(student): > a:=1;b:=-2;c:=1/4;d:=-5; > print(`khao sat ham so y=a*x^3+b*x^2+c*x+d`); > Y:= a*x^3+b*x^2+c*x+d; > print(`tap xac dinh cua ham so la R;`); > dy/dx=factor(simplify(diff(Y,x))); > print(`giai phuong trinhY'=0`); > solve(diff(Y,x)=0,{x}); > print(`ham so dong bien tren khoang`); > solve(diff(Y,x)>0); > print(`ham so nghich bien tren khoang`); > solve(diff(Y,x) print(`tim cac gia tri cuc tri dia phuong`); > Ymin_max:=extrema(Y,{},x); > print(`tinh dao ham bac hai cua ham so`); > z:=(simplify(diff(Y,x$2))); > print(`diem uon cua thi ham so`); > solve({z=0,Y=y},{x,y}); > print(`tim giao diem voi truc tung`); > student[intercept](y=Y,x=0,{x,y}); > print(`tim giao diem voi truc hoanh`); > student[intercept](y=Y,y=0,{x,y}); l > print(`do thi ham so co dang sau`); > plot(Y,x=-6 6,color=red); b) y = ax4 + bx2 +c, vi a, b, c nhp t bn phớm ax + b cx + d c) y = , vi a, b, c, d nhp t bn phớm restart;with(student); > a:=1;b:=-2;c:=2;d:=4; > Y:=(a*x+b)/(c*x+d); > print(`tap xac dinh cua ham so la: D=R`,{solve(denom(Y)=0,x)}); > print(`tinh dao ham cap cua ham so:`); > dy/dx=simplify(diff(Y,x)); > print(`giai phuong trinh Y'=0`); > solve(diff(Y,x)=0,{x}); > print(`ham so dong bien tren khoang:`); > solve(diff(Y,x)>0); > print(`ham so nghich bien tren khoang:`); > solve(diff(Y,x) print(`tim cuc tri dia phuong:`); > Ymin_max=extrema(Y,{},x); > print(`tim cac duong tiem can:`); > A:=limit(Y/x,x=infinity); > B:=limit(Y-A*x,x=infinity); > ms:=solve(denom(Y)=0,x); > if A=0 then print(`do thi ham so co duong tiem can ngang la Y=`,B); > else if Ainfinity or A-infinity then print(`do thi ham so co duong tiem can xien la Y=`,A*x+B);fi;fi; > print(`do thi ham so co duong tiem can dung la X=`,ms); > print(`tim dao ham cap cua ham so:`); > z:=simplify(diff(Y,x$2)); > print(`tim diem uon cua thi ham so:`); > solve({z=0,Y=y},{x,y}); > print(`do thi ham so giao voi truc tung tai:`); > student[intercept](Y=y,x=0,{x,y}); > print(`do thi ham so giao voi truc hoanh tai:`); > student[intercept](Y=y,y=0,{x,y}); > print(`do thi ham so co dang:`); > plot(Y,x=-10 10,color=red); y= ax2 + bx + c dx + e d) , vi a, b, c, d, e nhp t bn phớm Trong ú cn bit cỏc cõu lnh: tớnh o hm; gii phng trỡnh, bt phng trỡnh; tỡm cỏc im cc tr; tỡm tim cn; xỏc nh cỏc khong ng bin, nghch bin; xỏc nh cỏc khong li, lừm; v th hm s; tỡm giao im ca th vi trc tung, trc honh 12 S dng phn mm maple, gii cỏc bi toỏn hỡnh hc phng sau: 1) Cho tam giỏc ABC, bit A(1; 4), B(3; -1) v C(6; 2) a) Lp phng trỡnh tng quỏt ca cỏc ng thng AB, BC, CA b) Lp phng trỡnh tng quỏt ca ng cao AH v trung tuyn AM point(A,1,4); point(B,3,-1); point(C,6,2); triangle(T,[A,B,C]); > Equation(l,[x,y]);Equation(l,[x,y]);C >line(l,[A,B]); Equation(l,[x,y]); > line(l1,[B,C]); Equation(l1,[x,y]); > line(l2,[C,A]); Equation(l2,[x,y]); > midpoint(M,A,B);coordinates(M); > line(AM,[A,M]); Equation(AM,[x,y]); > print(`phuong trinh tong quat cua duong cao AH la`); > PerpendicularLine(AH,A,l1); > Equation(AH,[x,y]); 2) Lp phng trỡnh ng trũn cỏc trng hp sau: a) Cú tõm O(-2; 3) v i qua im M(2; -3) b) Cú tõm O(-1; 2) v tip xỳc vi ng thng x 2y + = c) Cú ng kớnh AB vi A(1; 1) v B(7; 5) d) i qua ba im A(1; 2), B(5; 2) v C(1; -3) 3) Cho ba im A(4; 3), B(2; 7) v C(-3; -8) a) Tỡm ta ca trng tõm G v trc tõm H ca tam giỏc ABC b) Gi T l tõm ca ng trũn ngoi tip tam giỏc ABC Chng minh T, G, H thng hng c) Vit phng trỡnh ng trũn ngoi tip tam giỏc ABC 4) Cho tam giỏc ABC vi H l trc tõm Bit phng trỡnh ca cỏc ng thng AB, BH v AH ln lt l 4x + y - 12 = 0, 5x - 4y - 15 = v 2x + 2y - = Hóy vit phng trỡnh hai ng thng cha hai cnh cũn li v ng cao th ba [...]... A=0 then print(`do thi ham so co duong tiem can ngang la Y=`,B); > else if Ainfinity or A-infinity then print(`do thi ham so co duong tiem can xien la Y=`,A*x+B);fi;fi; > print(`do thi ham so co duong tiem can dung la X=`,ms); > print(`tim dao ham cap 2 cua ham so:`); > z:=simplify(diff(Y,x$2)); > print(`tim diem uon cua do thi ham so:`); > solve({z=0,Y=y},{x,y}); > print(`do thi ham so giao voi... dia phuong`); > Ymin_max:=extrema(Y,{},x); > print(`tinh dao ham bac hai cua ham so`); > z:=(simplify(diff(Y,x$2))); > print(`diem uon cua do thi ham so`); > solve({z=0,Y=y},{x,y}); > print(`tim giao diem voi truc tung`); > student[intercept](y=Y,x=0,{x,y}); > print(`tim giao diem voi truc hoanh`); > student[intercept](y=Y,y=0,{x,y}); l > print(`do thi ham so co dang sau`); > plot(Y,x=-6 6,color=red);... solve({z=0,Y=y},{x,y}); > print(`do thi ham so giao voi truc tung tai:`); > student[intercept](Y=y,x=0,{x,y}); > print(`do thi ham so giao voi truc hoanh tai:`); > student[intercept](Y=y,y=0,{x,y}); > print(`do thi ham so co dang:`); > plot(Y,x=-10 10,color=red); y= ax2 + bx + c dx + e d) , vi a, b, c, d, e nhp t bn phớm Trong ú cn bit cỏc cõu lnh: tớnh o hm; gii phng trỡnh, bt phng trỡnh; tỡm cỏc im cc tr; tỡm tim cn; xỏc... hc: a) Nu kớ hiu pk l s nguyờn t th k (k t s 3 tr i) thỡ mt s nguyờn t vi mi n n Fn = 2 2 + 1 An = p1 p2 pn 2 b) Vi mi s nguyờn dng n, ta luụn cú l mt s nguyờn t 2) Kim tra v trớ tng i ca hai ng thng trong mt phng 3) Gii h phng trỡnh bc nht hai n 11 S dng phn mm Maple vit chng trỡnh kho sỏt hm s a) y = ax3 + bx2 + cx + d, vi a, b, c, d nhp t bn phớm restart: with(student): > a:=1;b:=-2;c:=1/4;d:=-5;... phớm ax + b cx + d c) y = , vi a, b, c, d nhp t bn phớm restart;with(student); > a:=1;b:=-2;c:=2;d:=4; > Y:=(a*x+b)/(c*x+d); > print(`tap xac dinh cua ham so la: D=R`,{solve(denom(Y)=0,x)}); > print(`tinh dao ham cap 1 cua ham so:`); > dy/dx=simplify(diff(Y,x)); > print(`giai phuong trinh Y'=0`); > solve(diff(Y,x)=0,{x}); > print(`ham so dong bien tren khoang:`); > solve(diff(Y,x)>0); > print(`ham... midpoint(M,A,B);coordinates(M); > line(AM,[A,M]); Equation(AM,[x,y]); > print(`phuong trinh tong quat cua duong cao AH la`); > PerpendicularLine(AH,A,l1); > Equation(AH,[x,y]); 2) Lp phng trỡnh ng trũn trong cỏc trng hp sau: a) Cú tõm O(-2; 3) v i qua im M(2; -3) b) Cú tõm O(-1; 2) v tip xỳc vi ng thng x 2y + 7 = 0 c) Cú ng kớnh AB vi A(1; 1) v B(7; 5) d) i qua ba im A(1; 2), B(5; 2) v C(1; -3) 3) Cho ... C lờn BP Tỡm qu tớch ca E Dng thit din, tỡm hiu hỡnh dng thit din: Cho hỡnh lp phng ABCD.ABCD, im M di ng trờn tia AC; im N, K ln lt nm trờn on thng AB, AD Xỏc nh thit din to bi mt phng i qua... then print(`do thi ham so co duong tiem can ngang la Y=`,B); > else if Ainfinity or A-infinity then print(`do thi ham so co duong tiem can xien la Y=`,A*x+B);fi;fi; > print(`do thi ham so co... > print(`tim diem uon cua thi ham so:`); > solve({z=0,Y=y},{x,y}); > print(`do thi ham so giao voi truc tung tai:`); > student[intercept](Y=y,x=0,{x,y}); > print(`do thi ham so giao voi truc hoanh