JOURNAL OF SCIENCE OF HNUE Educational Sci„ 2009, Vol 54, No 8, pp 3 13 T A P L U Y E N C H O H O C S I N H H O A T D O N G L I E N TU''''DNG VA H U Y D O N G K I E N TRU''''C T R O N G Q U A T R I N H C H[.]
JOURNAL OF SCIENCE OF HNUE Educational Sci„ 2009, Vol 54, No 8, pp 3-13 T A P L U Y E N C H O H O C SINH HOAT D O N G LIEN TU'DNG VA H U Y D O N G K I E N T R U ' C T R O N G Q U A T R I N H C H I E M L I N K T R I THLfC T O A N H O C Nguyin HfJu Hau Trudng THPT Dong San - Thanh Hod Md dau Day Toan la day hoat dong Toan hpc [1;12], cho nen mot nhung yeu cau ciia day Toan la phai khai day dugc kha nang doc lap suy nghi va kham pha cua ngUdi hoc Kha nang van dung tot nhiing kien thiic da dugc hoc de giai quyet nhQng vin de mdi la mot nhQng nhiem vu quan ciia viec hoc Lien tudng va huy dpng la nhQng kha nang rat quan trpng can phai ren luyen cho HS Ngu cd kha nang lien tuong tdt thi dQng trUdc mot bai toan kho, hpc sinh se lien he vdi cac kien thQe lien quan co kha nang giai quyet van de NgUpc lai, neu kha nang lien tudng kem thi gap mot van de nidi HS thudng khong biet dat nd moi lien he vdi cac kign thQe da biet, vi thg each nhin vS,n de mang tinh cue bp, rdi rac, Toan hpc la mot he thdng cac kign thiic cd lien he mat thiet vdi 2.1 N o i d u n g n g h i e n ciJu K h a i niem hen tu'dng va vai t r o ciia hen tifdng diidi goc tam h Theo tQ dign Tiing Viet, lign tudng cd nghia la: "Nhan su vat, hien tUpng nao dd ma nghi din sU vat hien tUpng khac c6 lien quan" [10; 568] Ta CO thg higu khai quat vg thuyit lien tudng thong qua mot so luan digm chfnh sau: - Tam If dupe cau tQ cam giac Cac cau cao hpn nhu bigu tUpng, y nghi, tinh cam, la cai thQ hai, xuat hien nhd lign tudng cac cam giac Noi each khac, dudng hinh tam If ngudi la lien kgt cac cam giac va cac y tudng; - Dien kien dg hinh lien tUPng la sU gan gui ciia cac qua trinh tam li; - Su lien ket cac cam giac va y tucfng dg hinh y tudng mdi khong phai la su kit hpp gian don cac cam giac hoac cac y tudng da cd, ma la sU kit hpp phQc tap tao nhung y tudng mdi doi mang tinh nhay vpt; - Cac lien tudng bi qui dinh bdi su linh hoat ciia cac cam giac va cac y tudng phan dupc lign tuctng va t i n so nhdc lai ciia chiing kinh nghiem; Nguyen HQu Hau - Cac lien tudng dupc hinh theo mot so qui luat: Qui luat tuong tu Qui luat tuong can Qui luat nhan qua [6;33,34] Cac nha lien tuong hpc quan tam hon ca tdi toe dp va miic dp lien kit cac hinh anh, cac bigu tupng da cd Theo hp, cd bon loai lien tuong la: Lign tudng giong nhau, Lign tudng tUdng phan, Lien tudng gkn ke vi khong gian va thdi gian, va Lign tudng nhan qua, dd, Lign tudng nhan qua cd vai trd dac biet quan trpng [6;34l Theo Bill Van Hue, lign tudng dupc chia bdn loai: lien tudng gdn ve khong gian va thdi gian, lien tudng giong vi hinh thii hoac noi dung, lign tudng trai ngupc nhau, lign tUPng nhan qua Theo tac gia, lien tudng co vai trd rat cjuan trpng viec ghi nhd va nhd lai [3;69] Nha tam If hpc P A Sgvarev da nghign cQu ti mi nhung moi lien tudng khai quat dpc dao va vai tro ciia chiing day hpc Ong chi rang, nhung moi lign tudng khai quat bao gom ba kieu cP ban: nhung lien tudng dupc biin ddi mot nua, nhung lign tudng - biin thien va nhQng lien tUPng cu thg - biin thien [7; 136] K K Plantondv xem tU la mot qua trinh gom nhieu giai doan ki tiip nhau, cu thg la: xudt hien cac lien tUdng, sang Ipc cac lien tudng va hinh cac giathuyit [4;121] Theo tac gia Nguyin Ba Kim va Vu Duong Thuy: "Trong day hpc, cdn chu y ren luyen cho HS ki nang biin doi xudi va ngUpc chiiu mOt each song song, nhdm giup cho viec hinh cac lign tudng ngupc diin dong thdi vdi viec hinh cac lien tudng thuan" [5; 174] Nhu vay, co thg thay rang: Vai tro cua lign tUPng qua trinh tu rdt quan trpng, lien tuong ciing ddng vai tro quan trpng hoat dpng tu giai toan Dao Van Trung cho rdng "Lign tudng la doi canh ciia tu duy, la cdu noi giai quyit vdn de" [9;73] 2.2 Lien tif dng va huy d o n g kign thtJc t r o n g day h o c T o a n Khi di cap din su phan loai cac bai toan, G Polia quan niem: "Mot su phan loai tot phai chia cac bai toan nhung loai (kiiu, dang) cho mdi bai toan xac dinh trudc mot phuPng phap giai" Dua vao muc dfeh cua bai toan, dng chia cac loai bai toan hai loai: nhung bai toan tim tdi, nhung bai toan chQng minh.Vi mQc dp khd, d i cua bai toan, G Folia cho rdng "Khong de dang xet doan vg mQc dp kho ciia bai toan, lai cang khd hon nua xac lap gia tri giao due cua nd" [7;132] Giao vien (GV) nen nam dupc each phan loai dp khd ciia bai toan d i phuc vu cho viec giang day Trudc giai toan, khd khdng dinh dupc chdc chan rdng se dung nhung kiin thQe (dinh nghia, dinh If, menh de, qui tdc, cdng thQe, ) nao, trQ dd la bai toan da cd thuat giai Trong qua trinh giai mot bai toan cu t h i nao dd, le duong nhign khong can huy dOng din mpi kiin thQe ma ngudi giai da thu thap, tfch luy dupc tQ trudc Can huy dpng nhung kiin thQe nao, cdn xem xet din nhQng moi lien he nao phu thuOc vao kha nang chpn Ipc ciia ngudi giai toan Ngudi giai toan da tfch luy nhung tri thQe ay, gid day riit va van dung mot each thfch hpp d i giai toan G Tap luyen cho hoc sinh hoat dong lien tudng vd huy dong kien thiCc qud trinh Folia gpi viec nhd lai cd chpn Ipc cac tri thQe nhu vay la sU huy dpng, viec lam cho chung thich Qng vdi bai toan dang giai la sU td chQc [7;310, 311] Trudc bdt tay vao giai mot bai toan cu the, ngudi giai da tich luy dupe rdt nhiiu kiin thQe, nhung dirng kiin thQe nao cho phii hpp thi that khong di dang Tham chf niu cd kem theo chi din ve nhung kiin thiic cdn a[) dung thi bai toan ehua hdn da giai dupc bdi vi, tliQ nhdt, chua hdn HS da nhd dupc noi dung kiin thQe dd, thQ hai, tlii nhd dupc noi dung kiin thQe thi viec ap dung no cung khdng han la d i dang Mat khac, nipt bai toan c6 chi ddn doi dua cac em vao trang thai bi dpng kiin thQe, cdn bai toan khdng dupc chi ddn lai dg cac em co thi tu lua chpn kiin thQe giiip giai quyit van de bdng nhiiu eon dudng khac Tuy nhien, kho va di chi la nhung khai niem mang tinh chat tuPng doi, phu thupc vao nhung tinh hudng, thdi digm cu thi Cac kiin thQe ma HS linh hoi dupc la san pham ciia hoat dpng, la mot bai toan ma muon chiim linh thi HS phai trai qua nhung hoat dpng tuPng Qng Kha nang huy dpng kiin tliQc khong phai la diiu b i t biin, nd thay doi theo lupng tfch luy tri thQe cua hpc sinh va mQc dp luyen tap thao ciia cac em HS huy dong kien th'iic de gidi quyet tot vdn de cdn thuoc vdo khd ndng Men doi vdn de, biSn doi bdi todn kho, m,di, la ve cdc bdi todn dcln gidn hon, quen thuoc Hon md cdc em dd tiing gidi quyet Chang han, xet bai toan: Tim m, de pInMng trinh sa,u co nghiem thuc: \ / F ^ ^ + m V-v~+T = \fx'^~^~ (2.1) Cac bai toan eo chQa tham s6 la mot dang toan tu'Piig doi kli6 vdi HS Viec chuyin bai todn da cho vi bai toan tUOng dUPiig khong dupc HS y thQe diy dii, co nhiiu sai 1dm khac vi mat tinh toan, van dung dinh li, qui tdc, kha nang khai quat van d i ciia cac em bi han ehi Mot yiu to khac cung gay kho khan cho hpc sinh la khau kiim tra dap an tUPng doi phQc tap (nhiiu khdng thuc hien dupe) day chung toi chi hudng din HS dinh hudng tim ldi giai bai toan budc biin ddi tQ dd lign tudng, huy dOng kiin thQe di dua vi bai toan mdi tuong dUPng vdi bai toan da cho giiip HS co diiu kien giai dung bai toan GV: Diiu kien xac dinh cua bai toan la gi? Cd t h i biin d6i dupc bai toan vi dang quen thupc hdn khdng? HS vdi cau tra ldi mong dpi: Diiu kign xac dinh ciia phuong trinh la :c > 1, va phuong trinh tUPng ditpng vdi phUPng trinh: cd nghiem 't: > Nhin vao phupng trinh (2.2) da biit each giai bai toan chua? HS vdi cau tra ldi mong dpi la dat an phu t = ^ -, vdi niQi c > 1, thi dieu kien ciia t Ik < t < 1, ngUpc lai vdi mdi t ma < t < thi diu eho ;; > Va Nguyin HQu Hau bai toan da cho tuong duong vdi bai toan: Tim diiu kien cua m de phuang rn = 2t-3t^ trinh: (2.3) cd nghiem < f < Cung phai ndi thgm rdng bai toan nay, viec tim dung diiu kien ciia / dg chuyin vi bai toan tUPng dUPng (2.3) la khdng di, b i n hit HS chi tim dupe diiu kien t > mh khong tim dupc diiu kien t < I Di tim dung diiu kien ciia / lai doi hoi HS phai huy dpng kiin thQe biin ddi va danh gia Ta co: o < f = a^ < 1, V.c> V-r+l GV hoi tiip: TQ phuong trinh (2.3) da biit each giai tiip bai toan chua? HS: So nghigm cua phUPng trinh (2.3) bdng s6 giao digm cua dd thi y = m (cung phuong vdi true O.r) va p h i n dd thi y = g{t) = 2t - 3t'^ xet tren < f < TQ HS giai dung bai todn: Gia tri cua m thoa man bai toan la < rn < ~ o Nhu vay, hoat dpng lien tUPng va huy dpng kiin thQe r i t quan trpng qua trinh giai toan GV cin dac biet chu y phat triin hoat dpng cho HS, giup cac em cd k h i nang dpc lap giai quyit cac bai toan Hoat dong Hen tudng vd huy dong kiin thdc moi ngudi moi khac Khd ndng lien tudng ndy phu thudc vdo su nha.y cdm d khdu phdt hien vdn de Khdng ed kha ndng lien tuPng va huy dpng kiin thQe thi nang luc giai todn bi han chi, cai nhin vi bai toan thudng cue bp, rdi rac Tuy nhign, dQng trUdc mot bai toan cu thg, khdng nhdt thiit tdt cd kiin th-dc lien tudng vd huy ddng duac deu cd ich cho viec gidi bdi todn C i n chpn Ipc thong qua cac phep thu - sai de tiin tdi mot su lign tUOng huy dpng phu hpp Chang han, xet bai toan: Chicng minh rdng niu o.,b.,c la ddi cdc canh cua mot tam gide thi a^ -I- b^ -\- c^ < 2{ah + be + ca) Ta nhan thiy rdng bai toan cd de cap din mdi quan he giua cac canh ciia mot tam giac, ta hay huy dOng nhQng dinh If, tinh chit da biit vi quan he giQa cac canh cua mot tam giac: a>b-c a b'^ — 2bc -\- (? Do vai tro cua o, b, c la binh ddng nhu nhau, bdi vd}' sau co a^ > b'^ — 2bc.+c'^ cuiig cin phai rut kit qua tuong tU: 6^ > a^ — 2ac -I- c^, c^ > b'^ — 2ba + a^ Cong tQng vi cac BDT ta cd diiu phai chQng minh Bay gid thu quan sat (2.5), eung suy nghi nhu tren, d i lam xuit hign a'^, eo thi binh phuong hai ve hoac nhan hai vi vdi a Niu tiin hanh theo dudng binh phUPng hai vi, ta cd a^ < b"^ -\- 2hc -I- c^, mot each tUdng tu cung cd 6^ < ci^ -\- 2ac -\- c^, c^ < a^ -I- 2ab + b~ Cong tQng vi ta suy a^ -I- 6" -f c" > -~2{a,b-\rbc + ca) Day la BDT hien nhign dung nhUng khong phai la dieu cin phai chQng minh Niu tiin hanh theo eon dudng nhan ea hai vi vdi a, a^ < ah -H ac Vi vai trd a, 6, c binh ddng nhu ngn ta cung cin su dung not nhung b i t ddng thQe tuong tU: h^ < bc-\- ah, c^ < ca -^ cb Cong tQng vi eac BDT ta co a^ + b'^ + c^ < 2{ab + bc + ca.) 2.3 N h i i n g quan digm chi dao viec t a p luyen cho H S hoat d o n g lien tifdng va huy d o n g kien thiJc * Quan diem Trong qud trinh truyen thu tri, thdc Todn hoc cho HS, GV cdn quan tdm tap luyen nhdn dang, phdt hien cdc the hien khde nhau, tii dd nhdn manh khd ndng icng dung cua nd bang viec lua chon he thdng bdi tap de HS thdy dugc moi lien he gvBa cdc ndi dung Todn hoc Trong day hpc thi viec truyin thu eho HS kiin thQe cP ban la r i t quan trpng, dd la nhung dinh nghia, dinh If hay cac qui tdc Nhung diiu quan trpng la hpc sinh phai cd kha nang van dung chung qua trinh lam bai tap Vi thi GV cin minh hpa kiin thQe bdng cdc vi du cu t h i sinh dpng di HS nhan biit each thQe Qng dung nhQng kiin thQe dd Chdng han, day hpc dinh If Lagrange "Cho ham lign tuc trgn doan [a, 6], co dao ham trgn khoang (a, b) Ton tai mot so c G (a, b) cho /'(c) = ", GV se gpi y cho HS phat hien cac Qng dung khac cua dinh li l}ng dung 1: Su dung dinh If d i chQng minh b i t ddng thQe Ta c6, niu m < F'ic) < M, Vc E (a, b) thi m < ^H^LzfM < M ^ m.{b - a) < F{b) - F{a) < b— a M{b -a) Do dd, d i dp dung dupc kit qua trgn vao viec chQng minh BDT, dieu quan trpng la phai tun dupc ham s6 F{x) Vf du: Chiing minh rang ln(x -^ 1) < x vdi moi x > Cho HS viit lai BDT trgn di lam xuit hien ham so F{x) : ln(x -|- 1) - In < Nguyin HQu Hau fx + 1) - o i ^ - t i l ^ ! ^ < Khi do, HS se d i dang lien tudng din viec ^ ^ ^ (:r + 1) - xet ham so F{t) = hit lign tuc va cd dao ham tren [l,.-?: + 1], {x > 0) theo dinh If Lagrange ludn tdn tai c G (1, x -H 1) vdi a; > cho: ,, , ^^ F(.T+1)-F(1) (x -h 1) - 1 c Ta c6 < c < T -M suy ln(.i- + 1) = ln(:c+l) X '-"; A B,C la ba gde cua mot tam giac nen cac hang tu bieu tliQc ludn duong Cin danh gia tdng cac sd dupng theo chiiu " > " gpi cho cac em lign tUdng tdi BDT Cauchy Tuy nhign, dp dung true tiip thi diu "=" khdng xay day, cau hdi ciia GV nham giup HS huy dpng thgm kiin thQe q /q sin A -\- sin B -\- sin C < di xac dinh hudng giai Trong gid hpc giai quyit vin di, cae can hdi nhdm vao viec gdi lai cac tri thQe CO lign quan vdn tri thQe da dupc linh hpi trudc day ciia HS Cac can hdi cua GV cd tac dung thue ddy budc tim toi tri tliQc co lien quan de tim hudng giai quyit thfch hpp, loai trQ dupe nhung sai lech cd thg tren budc dudng giai quyit dung ddn HS dua diiu minh da biit vdo nhung mdi lign he thfch hpp * Quan diem 4- Can tao cho HS tlidi quen nhin nhdn mdt vdn de dudi nhieu goc khac qud trinh truyen thu tri thdc Mot bai toan eo t h i nhin dudi nhieu khfa canh khac va Qng vdi mdi each nhin c6 thg cho ta mot ldi giai khae Hon nua mpi each giai diu dua vao mot sd dac diim nao cua cae du kign, eho ngn viec tim nhiiu each giai luyen cho HS biit nhin nhan vin d i theo nhieu khfa canh khac nhau, diiu r i t bd fch cho vige phdt triin ndng luc tu Mot ddc diim tam If cua HS ciua trinh giai todn la: Tu ludn cd siic i Di ren luyen tu linh hoat, phd vd siic i cua tu ta cdn phdi tliuc liien cdc hoat ddng thdnh phdn sau: -f Huy dpng kiin thQe lign quan din gia thiit va kit ludn cua bai toan theo 11 Nguyin Huu Hdu cac he thdng kien thQe lien (pian khac nhau; -t Hoat dpng chuyin ddi ngdn ngQ mot iipi dung Toan hpc hoac ehuyen ddi ngdn ngQ sang ngon ngU khac day hpc cac tinh hudng dign hinh; -^ Biin ddi bai toan bai toan khac tUcJiig dudng (niQc dp tdng quat eo t h i khae nhau) \'i du: Gidi phurJng trinh luang gidc: eos'-^ X -h 2\/3 sin x cos:;; -|- sin" x = (2.8) Ta CO t h i dinh hudng nhu sau: ThQ nhit: Bdi vi, eos" r -f siii^ x = nen cd t h i dua phUdng trinh (2.8) vi dang phuong trinh ddng cip bdc hai ddi vdi sin x, cos r ThQ hai: Do sd hang thQ hai eo chQa sin x cos x nen ed t h i dua (2.8) vi phupng trinh bac nhit ddi vdi sin2;r,cos2x nhd edng thQe bae ThQ ba: Xem cos".;; -I- \ / s i n x c o s x -I- s i n - x = la phuong trinh bac hai ddi vdi i n la cos:r TQ phudng trinh dd ta co A' = 3siii'^,r 3sin".r -F = Nhu vdy, phan tfch di dinh hudng bai toan chung ta phai lien tudng giua eac pham vi khac nhau, lien tudng den tQng chi tiit bai toan G Polia cho ring "Nhd nghien cQu lien tiip tiUig chi tiet mot, bdng nhiiu each, cudi cung chung ta cung cci t h i nhin dupe toan bp vin di dudi mot anh sang hoan toan khae trUdc va do nit mot each cliQng minh nidi" |7; 94] Hudng d i n cho HS co t h i van dung kiin thQe tdng hpp, kiin thQe lign mon, thiit lap su lien tudng giua eae pham vi dg tim nhiiu each giai eho mot bai todn Vf du: Cho ba sd duang a, b c thoa mdn a > c b > c Cliiing minh rdng: y ^ - r ) + V ^ ( - C ) < v^6 GV ed the dinh hudng giup HS giai bai toan nhu sau: Dinh hudng 1: Thiy giao cd t h i thuyit trinh: Nhin vao ciu tao eac biiu thQe BDT ta t h i y \fc+{a - c) = ^/d s/c-{- [b c) • \/h Neu nhdn tQng vi eua ddng tliQc ta ditpe: y^c + (« — c).\/c f {b - c) = y/ab GV ed t h i ngn cho HS can hdi: Vdi phep biin ddi eiia eae ddng thQe tren gpi cho em lien tudng din BDT nao? Mot BDT r i t quen thupc? Chung ta mong dpi HS tra ldi rdng: Do la BDT Bunliiaeopxki Vdi lign tudng nhu trgn, HS ap dung BDT Bunhiaeopxki: >/^-y/{^ c) + V'cViJ^- c) < y/{c + (a - c)).{c f {b - c)) - v/^Ift Ta cd dieu i)liai chQng minh 12 Tap luyen cho hoc sinh hoat dong lien tudng vd huy dong kien thiic qua trinh Dinh hudng 2: Vdi vige dp dung BDT Biiiihiaeoi)xki ta co lien tudng din tich vd hudng cua hai vec to, biiu thQe tfch vo hudng cua hai vec td dupc biiu thi dudi dang tpa dp c) > fiv = \/c\fa - c ^That vay, dat it =- {^/cT- c, ^/c) Ii = {s/c.\/h\fc\fb — c va |v7||?7| =- ^/dh, mat khac ta ludn co u.v < \u\\f.'\ nen ta dupc y^cy^a - c-\y/c\/b~— c < Vdh Tuy thupc vao each nhin vd kha nang biin ddi bai todn ciia HS, tQ d(3 G\' c6 thg dinh hudng eho HS bdng each ehuyen ddi sang ngdn iigu hinh lipe ngon ngu lupng giac hoac cd t h i biin ddi tUPng dupiig, sU dung eong eu (hu) ham Ket luan Cin thiit phai tap luyen cho HS kha nang lien tUPng va huy dpng kiin thii'c cpia trinh chiim linh tri thQe Thue hien diiu la goj) i)hin phat triin tu toan hpc cho HS Tuy nhign, mudn dat dupe diiu cin phai lua chpn noi dung va phUPng phap day hpc mot each thich hpp TAI LIEU T H A M K H A O [1] A A Stoliar, 1969 Gido due hoc Todn hoc Nxb Giao due i\linsk(Tiing Nga) [2] M Alecxeep, V Onhisuc, M Gruglidc, V Zabontin X Veexcle, 1976 Phdt trien tu hoc sinh Nxb Giao due Ha Noi [3] Bui Vdn Hue, 2000 Gido trinh tdm li hoc Nxb Dai hpc Qudc gia Ha Noi >' [4] Pham Minh Hac, Pham Hoang Gia, Trin Trpng Thuy NgUA'in Quang Udn, 1992 Tdm li hoc NXIJ Gido d u e Hd Noi [5] Nguyin Ba Kim, Vu Dudng Thuy, 2001 Pliuang ptidp dq.y hoc man tod,n Nxb Giao due Ha Ndi [6] Phan Trpng Ngp, 2005 Da,y hoc vd pliuang phdp day hoc nhd trudng, Nxb Dai hpc Su pham [7] A V Ptrdvxki, 1982 Tdm li hoc h'ia tudi vd td,m li hoc ,sii ph(im, tap Nxb Gido due Ha Ndi [8] G Polia, 1995 Sdng tao todn hoc Nxb Giao due, Ha Noi[9] Dao Vdn Trung, 2001 Ldm thi ndo de hoc tot todn phd tfwng Nxb Dai hpc Qudc gia, Ha Noi [10] Tic die'n Tiing Viet 2005 Nxb Da Ndng vd Trung tdm TQ diin hpc, Ha Ndi - Da Ndng ABSTRACT Training students associating activities and mobilizing knowledge during making up and compiling m a t h e m a t i c s knowledge In this paper, we dealt with training students associating activities and mobilizing knowledge during making up and compiling preparing mathematics knowledge The main result is to indicate the role of associating activies and mobilizing mathematical knowledge The methodology of teaching and carrying out the main ideals to drill activities for students 13 ... giQa tri thQe sdn cd vdi cac kham pha dpc lap cua HS cpia trinh hpc tap Ndi each khac, theo quan diim eua Vugotxki, dd chfnh la qua trinh di chuyin tri thQe tQ "viing phat triin gin nhit" din trinh... Qng dung cua nd * Quan diem Cdn chuyen tri thUc day hoc vi vimg phdt trien gdn nhdt tap luyen cho HS hoat ddng lien tudng vd huy ddng kiin thiic Thuc tiin su pham cho thiy, d i phat hien vin di,... g quan digm chi dao viec t a p luyen cho H S hoat d o n g lien tifdng va huy d o n g kien thiJc * Quan diem Trong qud trinh truyen thu tri, thdc Todn hoc cho HS, GV cdn quan tdm tap luyen nhdn