Tich Phan Kep

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tnh tch phn

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<ul><li><p>CHNG II: TCH PHN BI0: MT S MT BC HAI THNG GP1: TCH PHN KPnh ngha v Cch tnhi bin trong tch phn kpng dng ca tch phn kp2: TCH PHN BI BAnh ngha v Cch tnhi bin trong tch phn bi bang dng ca tch phn bi ba </p></li><li><p>0. Mt s mt bc hai thng gpMt Ellipsoid:2. Cch gi tn mt: Vi phng trnh trn, ta cho x = 0, y = 0, z = 0 ta u nhn c giao tuyn ca mt vi 3 mt ta l cc ng Ellipse. 3. Cch v hnhV 3 giao tuyn ca S vi 3 mt ta Nu c 3 giao tuyn ca 1 mt cong S vi 3 mt ta hoc cc mt song song vi cc mt ta u l ellipse th ta s gi mt S l mt Ellipsoid</p></li><li><p>0. Mt s mt bc hai thng gp</p></li><li><p>0. Mt s mt bc hai thng gp</p></li><li><p>0. Mt s mt bc hai thng gpV mt ellipsoidTrong MatLab, v ellipsoid trn, ta dng lnh ellipsoid(a,b,c)</p></li><li><p>0. Mt s mt bc hai thng gp</p></li><li><p>0. Mt s mt bc hai thng gpII. Mt Paraboloid Elliptic:2. Cch gi tn mt:Vi phng trnh trn, ta cho x = 0, y = 0 th c 2 giao tuyn vi 2 mt ta l 2 ng Parabol v cho z=c, c&gt;0 ta c ng cn li l 1 ng Ellipse. Nu 2 trong 3 giao tuyn vi cc mt ta hoc cc mt song song vi cc mt ta l 2 Parabol, giao tuyn cn li l 1 Ellipse th ta gi mt S l Paraboloid Elliptic</p></li><li><p>0. MT S MT BC HAI THNG GPV ng parabol y2 = z trn mt phng x = 03. V hnh</p></li><li><p>0. MT S MT BC HAI THNG GPV ng ellipse x2+y2 = 1 trn mt phng z = 1</p></li><li><p>0. MT S MT BC HAI THNG GPV mt parabolid x2+y2 = z</p></li><li><p>0. MT S MT BC HAI THNG GPV thm ng parabol x2 = z trn mt phng y = 0</p></li><li><p>0. MT S MT BC HAI THNG GPIII. Mt Tr bc 2:nh ngha mt tr bc 2:Mt tr bc 2 l mt to bi cc ng thng song song vi 1 phng c nh v ta ln 1 ng cong bc 2 c nh. Cc ng thng gi l cc ng sinh ca mt tr, ng cong c nh gi l ng chun ca mt tr.</p></li><li><p>Thng thng, ta s ch gp cc mt tr c ng sinh song song vi 1 trong 3 trc ta . 0. MT S MT BC HAI THNG GPMt tr song song vi trc no th phng trnh mt s thiu bin , cn phng trnh bc 2 cha 2 bin cn li l phng trnh ng chun ca mt tr trong mt ta tng ng v ta gi tn mt tr theo tn ca ng chun</p></li><li><p>0. MT S MT BC HAI THNG GPV ng trn x2+y2=1, trn mt z=0Mt tr to bi cc ng thng song song vi Oz v ta ln ng trn trnV d: Mt x2+y2 = 1Phng trnh khng cha z nn n biu din mt tr ng sinh song song vi trc Oz, ng chun l ng trn x2+y2=1 trong mt phng z = 0 v ta gi y l mt tr trn xoay theo tn ca ng chun</p></li><li><p>0. MT S MT BC HAI THNG GPTrong MatLab, v tr trn xoay c th dng lnh cylinder</p></li><li><p>0. MT S MT BC HAI THNG GPV d : Mt z=x2Phng trnh khng cha y nn n biu din mt tr song song vi trc Oy, ng chun l parabol z=x2 trn mt phng y=0 nn ta gi y l mt tr parabolV parabol z=x2 trong mt phng y=0V mt tr c ng sinh song song vi trc Oy, ta ln ng chun l parabol z=x2 trn</p></li><li><p>0. MT S MT BC HAI THNG GPIV. Mt nn bc 2 :Mt nn bc 2 l mt to bi cc ng thng i qua 1 im c nh v ta ln 1 ng cong c nh. Cc ng thng gi l cc ng sinh ca mt nn, ng cong c nh gi l ng chun ca mt nn v im c nh gi l nh ca nnV d: Mt nn x2+y2=z2Ct dc mt nn bi cc mt x=0 hoc y=0 ta c 2 ng thng cng i qua gc ta O, ct ngang bi mt z = c v z = -c , c ty , ta c giao tuyn l 2 ng trn tm ti (0,0,c) v (0,0,-c) bn knh bng c</p></li><li><p>0. MT S MT BC HAI THNG GPV giao tuyn x2+y2=1, z=1V giao tuyn x2=z2, y=0V mt nn x2+y2=z2, ly phn z &gt; 0 </p></li><li><p>0. MT S MT BC HAI THNG GPV d: Nhn dng v v mt bc 2 sau z = x2+y2-2xGii: Ta ln lt cho x = 0, y = 0, z = 0 tm 3 giao tuyn ca mt cho vi 3 mt ta x = 0 : z = y2 l phng trnh paraboly = 0 : z = x2-2x l phng trnh parabolz = 0 : 0 = x2+y2-2x l pt ng trn (ellipse)Suy ra mt cho l mt Paraboloid EllipticNHN DNG</p></li><li><p>0. MT S MT BC HAI THNG GPV HNH:V 2 giao tuyn vi 2 mt z = 0, y = 0x=0:.1:2;z=0*x;y=sqrt(2*x-x.^2);plot3(x,y,z)hold ony=-sqrt(2*x-x.^2);plot3(x,y,z)Ta c giao tuyn vi z=0</p></li><li><p>0. MT S MT BC HAI THNG GPhold ony=-2:.2:2;x=1+0*y;z=-1+y.^2;plot3(x,y,z)</p></li><li><p>0. MT S MT BC HAI THNG GPV mt&gt;&gt; [r p]=meshgrid(linspace(0,1,20),linspace(0,2*pi,20));&gt;&gt; mesh(r.*cos(p),r.*sin(p),r.^2)</p></li><li><p>0. MT S MT BC HAI THNG GPV d: Nhn dng v v mt bc 2 sau x2+y2+z2-2z=0Gii: Ta ln lt cho x = 0, y = 0, z = 0 tm 3 giao tuyn ca mt cho vi 3 mt ta NHN DNGx = 0 : y2+z2-2z=0 l pt ng trn (ellipse)y = 0 : x2+z2-2z=0 l pt ng trn (ellipse)z = 0 : 0 = x2+y2l pt ng trn (ellipse)Suy ra mt cho l mt Ellipsoid</p></li><li><p>0. MT S MT BC HAI THNG GP&gt;&gt; theta=linspace(0,pi,20);&gt;&gt; phi=linspace(0,2*pi,20);&gt;&gt; [t p]=meshgrid(theta,phi); &gt;&gt; mesh(sin(t).*cos(p),sin(t).*sin(p),1+cos(t))</p></li><li><p>0. MT S MT BC HAI THNG GPV d: Nhn dng v v mt bc 2 sau y2-z2+2y=0Gii: Pt khng cha x nn n biu din mt tr ng sinh song song vi trc OxNHN DNGTrong mp x = 0 : y2 - z2 + 2y = 0 l pt ng hyperbol tc l ng chun l ng hyperbol.Suy ra mt cho l mt Tr Hyperbol</p></li><li>&gt;&gt; [x y1] =meshgrid(linspace(-1,1,20),linspace(-4,-2,20));&gt;&gt; z1=sqrt(y1.^2+2*y1);&gt;&gt; mesh(x,y1,z1)&gt;&gt; hold on&gt;&gt; mesh(x,y1,-z1)0. MT S MT BC HAI THNG GPTng t, ta v na cn li ng vi 0</li><li><p>0. MT S MT BC HAI THNG GPV d: Nhn dng v v cc mt bc 2 sau:y2-z2+2x2=0x2+2x+2z2-3y=0xy=z21. 2 trong 3 giao tuyn l 2 cp t, giao tuyn th 3 l ellipse nn ta c mt nn ellipse2. 2 trong 3 giao tuyn l 2 parabol, giao tuyn th 3 l ellipse nn ta c mt Paraboloid elliptic3. t x=u+v, y=u-v th ta c ptu2-v2=z2 u2=v2+z2 l pt ca mt nn </p></li><li><p>1: Tch phn kp nh ngha v cch tnhSau y, ta s v hnh khi D l hnh ch nht</p></li><li><p>1: Tch phn kp nh ngha v cch tnhChia min D thnh n phn ty Dij bi cc ng thng song song vi 2 trc Ox, Oy. Ti mi min Dij ly 1 im M(xi,yj) ty Dij</p></li><li><p> 1: Tch phn kp nh ngha v cch tnhTh tch cc hnh hp nh vi y di l Dij, trn l phn mt z=f(x,y) s c tnh xp x vi hnh hp ch nht y l Dij, chiu cao l f(xi,yj). </p></li><li><p> 1: Tch phn kp nh ngha v cch tnhKhi , vt th ban u c th tch xp x vi tng th tch cc hnh hp ch nht nh xp lin tip nhau</p></li><li><p> Chia min D thnh n phn khng dm ln nhau l D1, D2, D3, (cc phn khng c phn chung) tng ng c din tch l S1, S2, S3, Trn mi min Dk ta ly 1 im Mk(xk,yk) ty . Hin nhin tng trn ph thuc vo cch chia min D v cch ly im Mknh ngha tch phn kp : Cho hm f(x,y) xc nh trong min ng, b chn D 1: Tch phn kp nh ngha v cch tnh</p></li><li><p> Cho n sao cho max{d(D)} 0 (d(D) l k hiu ng knh ca min D tc l khong cch ln nht gia 2 im bt k thuc D)Nu khi y tng Sn tin n gii hn hu hn S khng ph thuc vo cch chia min D cng nh cch ly im Mk th gii hn S c gi l tch phn kp ca hm f(x,y) trn min D v k hiu l Hm f(x,y) c gi l hm di du tch phn, D l min ly tch phn, ds l yu t din tch. Khi y, ta ni hm f(x,y) kh tch trn min D 1: Tch phn kp nh ngha v cch tnh</p></li><li><p> Ch : Nu f(x,y) kh tch trn D th ta c th chia D bi cc ng thng song song vi cc trc ta . Lc Dij s l hnh ch nht vi cc cnh l xi, yj nn Sij = xi. yj v ds c thay bi dxdy. V vy, ta thng dng k hiu 1: Tch phn kp nh ngha v cch tnh</p></li><li><p>iu kin kh tch : nh ngha ng cong trn : ng cong C c phng trnh tham s y = y(t), x = x(t) c gi l trn nu cc o hm x(t), y(t) lin tc v khng ng thi bng 0. ng cong C c gi l trn tng khc nu c th chia n thnh hu hn cc cung trn.nh l: Hm lin tc trn 1 min ng, b chn v c bin trn tng khc th kh tch trn min . 1: Tch phn kp nh ngha v cch tnhTnh cht : Cho f(x,y), g(x,y) l cc hm kh tch trn D2.</p></li><li><p>Tnh cht 1: Tch phn kp nh ngha v cch tnh3.</p></li><li><p>nh l: (V gi tr trung bnh ) 1: Tch phn kp nh ngha v cch tnh</p></li><li><p> V d : Cho vt th c gii hn trn bi mt bc hai f(x,y) = 16 x2 2y2, gii hn di bi hnh vung D = [0,2]x[0,2] v gii hn xung quanh bi 4 mt phng x=0, x=2, y=0, y=2. c lng th tch ca vt th trong cc trng hp sau : Chia D thnh 4 phn bng nhau; Chia D thnh 16 phn bng nhau; Chia D thnh 64 phn bng nhau;Chia D thnh 256 phn bng nhau;Tnh th tch vt th1: Tch phn kp nh ngha v cch tnh</p></li><li><p> 1: Tch phn kp nh ngha v cch tnh </p></li><li><p>1: Tch phn kp nh ngha v cch tnh</p></li><li><p>1: Tch phn kp nh ngha v cch tnh</p></li><li><p>1: Tch phn kp nh ngha v cch tnh</p></li><li><p>nh l Fubini: (Cch tnh tch phn kp) Cho hm f(x,y) lin tc trn min ng v b chn D 1: Tch phn kp nh ngha v cch tnhab</p></li><li><p> 1: Tch phn kp nh ngha v cch tnhx=x1(y)</p></li><li><p> Gii cu e) Tnh th tch ca vt th.22=48</p></li><li><p>Ta i tch phn ny bng 2 cch Cch 1 : Chiu min D xung trc Ox ta c on [1,4]i theo trc Oy t di ln1: Tch phn kp nh ngha v cch tnh</p></li><li><p> Cch 2 : Chiu min D xung trc Oy ta c on [-1,3]A(1,-1)B(1,3)C(4,0)-13i theo trc Ox t tri sang th khng ging nh trn, ta s gp 2 ng BC v AC. Do , ta s chia min D thnh 2 phn D1 v D2D1D2x=-y+41: Tch phn kp nh ngha v cch tnh</p></li><li><p> 1: Tch phn kp nh ngha v cch tnh</p></li><li><p>1: Tch phn kp nh ngha v cch tnhTa cn c th xc nh cn ca tch phn trn m khng cn v hnh nh sau:x = -2, x = 1x2+x-2 = 0Vy ta c -2 x 1, tc l ta ly trong khong 2 nghim ca tam thc f(x) = x2+x-2 nn ta c bt ng thc:x2+x-2 0x 2-x2Tc l, vi x nm trong khong (-2,1) th ng thng y=x nm di ng parabol y = 2-x2. Vy ta cng c </p></li><li><p>D1D2D3D4Min D c chia thnh 4 phn 1: Tch phn kp nh ngha v cch tnhV d : Tnh tch phn trong D l min gii hn bi : /4max{|x|,|y|} /2</p></li><li><p>Ta cn c th tnh tch phn ny bng cch tnh tch phn trn hnh vung ln tr tch phn trn hnh vung nh1: Tch phn kp nh ngha v cch tnhTng t, ta tnh cho 3 tch phn trn 3 min cn li.</p></li><li><p>V d: Tnh tch phn kp D l min gii hn bi -1x1, 0y1D11: Tch phn kp nh ngha v cch tnhD l min gii hn bi -1x1, 0y1V d: Tnh tch phn kp D l min gii hn bi -1x1, 0y1</p></li><li><p>Nu ch nhn vo min ly tch phn ny th ta chiu D xung trc no cng nh nhau. 1: Tch phn kp nh ngha v cch tnhTuy nhin, hm di du tch phn s buc ta phi chiu D xung trc Oy</p></li><li><p>Chiu min D va v xung trc Ox1: Tch phn kp nh ngha v cch tnhTa v min ly tch phn Ta thy phi chia D thnh 2 phn D1 v D2</p></li><li><p>Nhc li v ta cc im M c ta l (x,y) trong ta Descartes. Khi , mi lin h gia x, y v r, l 1: Tch phn kp i bin sang ta cct : </p></li><li><p>V d: i cc phng trnh sau sang ta cci sang ta cc m rng bng cch t : Th ta c pt r = 1 1: Tch phn kp i bin sang ta cc1. (x-a)2 + y2 = a2 x2 + y2 = 2ax r = 2acos3. x = 3 rcos = 3x=a.r.cos, y = b.r.sin</p></li><li><p>Cng thc i bin sang ta ccTrong Thng thng, ta s i tch phn kp sang ta cc nu min ly tch phn kp l 1 phn hnh trn hoc ellipse 1: Tch phn kp i bin sang ta cc</p></li><li><p>1: Tch phn kp i bin sang ta cc xc nh cn ca tch phn theo , ta qut t di ln theo ngc chiu kim ng h bi cc tia mu . Ta c i t 0 n /2</p></li><li><p>1: Tch phn kp i bin sang ta ccNu ch gp 1 ng nh trong v d ny th cn di ta s ly l 0, cn trn l pt ng trn sau khi sang ta cc: r = 2cosVy :</p></li><li><p>1: Tch phn kp i bin sang ta cci t trong gc ta ra ch gp 1 ng nn 0 r ay=3x y/x = 3 = /3</p></li><li><p>1: Tch phn kp i bin sang ta ccy &gt; 0, x+y=0 = 3/4 Suy ra : 3/4 x2+y2 = 2y r = 2sinSuy ra : 0 r 2sin</p></li><li><p>V d : Tnh tch phnTrong D gii hn bi : 1: Tch phn kp i bin sang ta ccV d : Tnh tch phn2x x2+y2 4x 2cos r 4cosy l trng hp ta c th khng cn v hnh cng ly c cn tch phn</p></li><li><p>V d : Tnh tch phn Trong D gii hn bi 211-11: Tch phn kp i bin sang ta cc</p></li><li><p>211-1Do vy, ta i tch phn ny bng cch di trc ta tm hnh trn l (0,0), sau mi i sang ta cc.Thc hin 2 vic trn bng 1 php i bin sang ta cc m rng nh sau: t1: Tch phn kp i bin sang ta cc</p></li><li><p>Khi , min D gii hn bi Vy :1: Tch phn kp i bin sang ta cc</p></li><li><p>V d : Tnh tch phn Trong D gii hn bi abTa i bin sang ta cc m rng bng cch t Th D gii hn bi 1: Tch phn kp i bin sang ta cc</p></li><li><p>ng dng hnh hc ca tch phn kp 1. Din tch hnh phng: Din tch min D trong mt phng Oxy c tnh bi 2. Th tch vt th gii hn trn bi mt gii hn di bi mt v gii hn xung quanh bi mt tr song song vi trc Oz c ng chun l bin min D c tnh bi: 1: Tch phn kp ng Dng</p></li><li><p>V d 1: Tnh din tch min phng D gii hn bi y2+2y-3x+1=0, 3x-3y-7=0Ta tm giao im 2 ng cong bng cch kh x t 2 ptTc l chiu min D xung trc Oy c on [-2,3]Khi -2 y 3, suy ngc li phng trnh (1) ta s c y2 + 2y + 1 3y + 7 Vy :1: Tch phn kp ng dng</p></li><li><p>1: Tch phn kp ng dngTrc tin, ta tm giao im cos = 3/2 = /6 , = -/6Vy : </p></li><li><p>Khi vt th gii hn ch bi 2 mt th ta tm hnh chiu D ca n xung mt phng z=0 bng cch kh z t 2 phng trnh 2 mt1: Tch phn kp ng dngFile: VD3_tpkep3.m</p></li><li><p>Vi bt ng thc hnh trn, ta thay ngc ln phng trnh (1) c Tc l mt nn l mt gii hn di, mt cu l mt gii hn trn ca vt th. Vy : 111: Tch phn kp ng dng</p></li><li><p>V d 4: Tnh th tch vt th gii hn bi x2 + y2 = 4, y2 = 2z, z=01: Tch phn kp ng dngTrong 3 mt to nn vt th, c 1 hnh tr kn (ng chun l ng cong kn) x2+y2=4 song song vi trc Oz (pt khng cha z) nn hnh chiu ca n xung mt z = 0 l hnh trn, tc l ta c min ly tch phn D: x2 + y2 4. D dng thy bt ng thc 0 .y2 , tc l mt z = 0 pha di v z = .y2 pha trnTa cn li 2 mt v phi xc nh mt no nm trn, mt no nm di c hm di du tch phn</p></li><li><p>1: Tch phn kp ng dngSuy ra hm di du tch phn l : Vy th tch cn tnh l : File: VD4_tpkep3.m</p></li><li><p>Ta s tm hnh chiu ca V xung mt phng Oxy da trn cc pt khng cha z tc l cc hnh tr c ng sinh song song vi trc OzTrong 4 mt cho c 2 mt tr (phng trnh khng cha z) cng song song vi Oz l y=1, y = x2Hai mt tr c 2 ng chun to thnh min D ng trong mt Oxy1: Tch phn kp ng dng</p></li><li><p>Vi 2 mt cn li hin nhin ta c 0 x2+y2 tc l f(x,y) = x2+y21: Tch phn kp ng dngy=x2y=1z=x2+y2File: VD5_tpkep3.m</p></li><li><p>Cc mt cng song song vi Oz (phng trnh khng cha z) l y = 0, 3x+y = 4, 3/2x+y = 4. 1: Tch phn kp ng dngV 3 t ny trong mp Oxy ta c ABC nn hnh chiu ca V xung mp Oxy l Dxy: ABC </p></li><li><p>1: Tch phn kp ng dngCn 2 mt m pt cha z l gip ta c hm di du tp nh b..t hin nhin:</p></li><li><p>1: Tch phn kp ng dng</p></li><li><p>V d 7 : Tnh th tch vt th gii hn bi : y = 0, z = 0, z = a x - y, 3x + y = a, 3/2x + y = aTrong 5 pt cho c 3 pt khng cha z tng ng vi 3 mp cng song song vi trc Oz1: Tch phn kp ng dng3 t ny gip ta c hnh chiu xung mt phng Oxy l ABC = Min DCn li 2 mt c pt cha z, ta s tm cch xc nh mt nm trn, nm di c hm di du tch phn </p></li><li><p>R rng, trn hnh v ta c ABC nm pha di ng thng a-x-y=0 tc l trong min D ta c bt ng thc 0 a-x-y. Suy ra hm di du tch phn l f(x,y) = a-x-y1: Tch phn kp ng dngTa i so snh z= a-x-y vi z= 0 bng cch v thm ng a-x-y=0 trong mt phng z=0 ang xtVy </p></li><li><p>1: Tch phn kp ng dngTa xoay trc Oy thng ng, ta s thy vt th chnh l hnh chp t gic, th tch bng 1/3 chiu cao nhn din tch y</p></li><li><p>1: Tch phn kp ng dngTa cng bt u tm hnh chiu ca vt th xung mt z = 0 bng cch ch ra cc mt tr vi pt khng cha zVi v d ny, ta ch c 2 mt l y=x v y = 3x tng ng vi 2 ng thng khng cho ta min ng D. V vy, ta s tm thm giao tuyn ca cc mt cn li vi mt z=0</p></li><li><p>1: Tch phn kp ng dngCho z = 0 v thay vo phng trnh Paraboloid ta c x2+y2 =1, tc l giao tuyn ca mt Paraboloid vi mt ta z = 0 l ng trn. 1 phn ng trn s Y KN phn cn m gia 2 ng thng trn.T suy ra, D l 1 phn hnh trn x2+y21 nm gia 2 ng thng vi x, y 0 Vi mi (x,y) thuc D, ta u c : 0 1-x2-y2 tc l mt phng z = 0 nm di v paraboloid z = 1-x2-y2 nm trn</p></li><li><p>1: Tch phn kp ng dngVy: V min ly tch phn l hnh trn nn ta s i sang ta cc bng cch t x=rcos, y=rsin Khi , ta c</p></li><li><p>Hai mt tr cng song song vi trc Ox l 1: Tch phn kp ng dngV bng din tch hnh trn ln tr din tch hnh trn nh</p></li><li><p>1: Tch phn kp ng dng</p></li><li><p>C. Din tch mt cong : Din tch phn mt cong S c phng trnh z = f(x,y) v c hnh chiu xung mt phng Oxy l min D c tnh biNh vy, tnh th tch vt th hoc tnh din tch 1 phn mt cong th trc tin ta phi xc nh c hnh chiu D ca vt th hoc phn mt cong cn tnh xung 1 trong 3 mt ta Oxy, Oyz, OzxVi mt cong cn tnh din tch, ta phi vit li pt mt bng cch vit 1 bin theo 2 bin cn li tu vo vic ta tm hnh chiu xung mp to no.1: Tch phn kp ng dng</p></li><li><p>1: Tch phn kp ng dng tnh din tch mt cong S nh tch phn kp, ta phi xc nh c hnh chiu D ca mt cong xung 1 trong 3 mt ta .Vi v d ny, ta s tm hnh chiu ca S xung mt z=0 bng cch kh z t 2 phng trnh choz2 = 4-x2-y2 = x2+y2 x2+y2 = 2 T phng trnh trn, ta c hnh chiu ca S xung mt z = 0 l hnh trn Dxy : x2+y2 2Sau , v tm hnh chiu xung mt z = 0 nn ta s tnh z=f(x,y) t phng trnh mt S</p></li><li><p>V mt S nm pha trn mt nn tc l z 0 nn ta ly 1: Tch phn kp ng dngVy:</p></li><li><p>2 mt phng cho u song song vi trc Ox (Pt khng cha x) nn ta s tm hnh chiu ca S xung mt phng x = 0Chiu 2 mt phng xung mt x = 0 ta c 2 ng thng cng i qua gc ta tc l cha c min...</p></li></ul>