chủ Đề tự chọn toán 10

Trng THPT Vinh Lc T Ton

Ch t chn Ton 10

Ch 1HM S V TH (3 tit)

I.MC TIU: Qua bi hc HS cn:

1. V kin thc:

Nm c kin thc v hm s v th: Khi nim, tp xc nh, tnh n iu ca hm s, v c th v da vo th lp bng bin thin ca hm s, xc nh c tnh chn (l) ca hm s.

2. V k nng:

- Tm c tp xc nh, bit cch kho st s bin thin v v c th ca mt hm s y = ax + b, hm s y = v th ca hm s y = ax2 + bx + c.

Bit xc nh cc hm s y = ax + b v y = ax2 + bx + c.

3. V t duy: Rn luyn k nng gii ton, t duy lgic, bit quy l v quen.4. V thi : Cn thn, chnh xc.II. CHUN B CA GV V HS:

*i vi HS: Nm vng kin thc v th v hm s, son bi v lm bi tp trc khi n lp.*i vi GV: Gio n, bi tp trc nghim, phiu hc tp,

IV. TIN TRNH BI HC V CC HOT NG:

(c chia thnh 3 tit)

Tit 1: n Tp kin thc v hm s v th v cc phng php gii cc dng ton c bn.Tit 2: Rn luyn k nng gii ton.

Tit 3: Rn luyn k nng gii ton v luyn tp.---------------(o0o(-----------------

Tit 1: N TP KIM THC V HM S V TH1) n nh lp, chia lp thnh 6 nhm (hoc nhiu hn ty thuc s lng HS trong lp)

2) Kim tra kin thc c:

GV: Nh ta bit, mt hm s f xc nh trn tp D

EMBED Equation.DSMT4 l mt quy tc t tng ng mi s x thuc D vi mt v ch mt s f(x). S y = f(x) gi l gi tr ca hm s f ti x, x gi l bin s ca hm s f. Tp D gi l tp xc nh (hay min xc nh) ca hm s f.

GV: Nu cc cu hi sau n kin thc c:

-Vy tp xc nh D ca hm s f l g?

- th ca hm s y = f(x) xc nh trn D l g?- Nu ta cho mt hm s y = f(x) xc nh trn khong (a ; b) th:

+ Hm s y = f(x) gi l ng bin (hay tng) trn D th n phi tha mn iu kin g?

+ Tng t i vi trng hp hm s nghch bin (hay gim).

-Nu trng hp chn (l) ca hm s.

GV: Nu phng php tm tp xc nh ca hm s v ly cc v d minh ha

*Dng a thc: f(x) = axn + bxn-1+ + cx + d

Hm s y = f(x) xc nh vi mi x

*Dng phn thc: f(x) =

iu kin hm s xc nh: B 0*p dng:

TGHot ng ca GVHot ng ca HSNi dung

GV:Ly v d p dng

GV: Cho hc sinh tho lun theo nhm v gi 2 HS trnh by li gii.

GV: Gi HS nhn xt, b sung.

GV: Nhn xt, b sung v cho im.HS: Suy ngh trnh by li gii

KQ: a) Tp xc nh D=

b) Tp xc nh:

D=

HS: Nhn xt v b sung sai st(nu c)V d1: Tm tp xc nh ca cc hm s:

a)y = 4x2- 3x +2

b)y =

*Kho st s bin thin ca mt hm s.

GV: xt s bin thin ca mt hm s ta phi lm th no?

HS; Suy ngh v tr li cu hi

GV: Nu phng php xt s bin thin ca hm s y = f(x) trong khong (a; b) c tin hnh nh sau:Ly x1, x2 ty thuc khong (a; b), vi x1 x2.

Lp t s . Nu t s dng th hm s ng bin, ngc li nghch bin.

*p dng:TGHot ng ca GVHot ng ca HSNi dung

GV: Xem phng php v suy ngh gii cc bi tp sau:

GV: Yu cu HS nhm l suy ngh gii cu a), nhm chn gii cu b)

GV: Gi HS i din hai nhm ln bng trnh by li gii ca nhm mnh.

GV: Gi HS nhm khc nhn xt b sung.GV: B sung thiu st (nu c) v cho im.

*Hm s chn, hm s l:

GV: Mt hm s y = f(x) xc nh trn D gi l hm chn (l) khi n phi tha mn iu kin g?

GV: Nu bi tp p dng v hng dn gii cu a), cc cu b) c) d) e) yu cu hc sinh suy ngh lm xem nh bi tpHS: Suy ngh v trnh by li giiHS: i din nhm trnh by li gii:

a)Tp xc nh: D =

x1, x2 , x1x2, ta c:

=

=x12+x1x2+x22+3

=

Vy >0 vi mi x1, x2 thuc D, x1 x2. Do hm s ng bin trn ton trc s.

b)KQ: Hm s lun nghch bin trn (-;2) v (2;+).

Hm s y = f(x) xc nh trn D c gi l hm chn nu:

Ngc li, gi l hm s l nu:HS: ch theo di bi

V d 2: Kho st s bin thin ca cc hm s sau trn tp xc nh ca chng:a) y = x3 + 3x +1;

b) y =

p dng: Xt tnh chn - l ca cc hm s sau:

a) y = 3x4+3x2 2

b) y = 2x3 5x

c) y = x;

d) y =

e) y =

*Bng bin thin ca th hm s:TGHot ng ca GVHot ng ca HSNi dung

GV: Cho hm s y = ax+b (a 0). Hy lp bng bin thin ca hm s trong 2 trng hp a>0 v a 0:x - +

y + 0

-

*TH a 0) th ta c cng thc ca th hm s thay i nh th no?

BI TP TRC NGHIM:Hy chn kt qu ng trong cc bi tp1 v 2 sau:

1. Cho hm s f(x) = .Tp xc nh ca hm s l:

(a)

(b);

(c);(d).

2. Cho hm s f(x) = . Tp xc nh ca hm s l:

(a)

(b)

(c)

(d)

3. Cho hm s f(x) = . Hy chn khng nh sai trong cc khng nh sau:

(a)im (1; 2) thuc th ca hm s;

(b)im (-1; 2) thuc th ca hm s;

(c)im (0; 0) thuc th ca hm s;

(d)im (4; 18) thuc th ca hm s .

4. Hy ch ra khng nh sai trong cc khng nh:

(a)Hm s y = x2 l hm s chn;

(b)Hm s y = l hm s chn;

(c)Hm s y = x2+1 l hm s chn;

(d)Hm s y =(x+1)2 l hm s chn.

5. Cho hm s f(x) = -2x2 + 1. Hy chn khng nh ng trong cc khng nh sau:

(a) Hm s ng bin trn ;

(b)Hm s nghch bin trn ;

(c)Hm s ng bin trn (0;+), nghch bin trn (-;0);

(d)Hm s ng bin trn (-;0), nghch bin trn (0;+).

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TIT 2: RN LUYN K NNG GII TON1. n nh lp, chia lp thnh 6 nhm.

2. Kim tra bi c:

Cu hi:

a) im M0(x0;y0) thuc th hm s y = f(x) khi v ch khi no?

b) Mt hm s y = f(x) xc nh trn D th hm s ng bin, nghch bin, chn,l khi no?

c)T nh i xng ca hm s chn - l nh th no?

d) Tnh tin mt th hm s y = f(x) song song vi cc trc ta trong mt phng Oxy. Khi tnh tin ln trn, xung di, qua phi, qua tri k n v (k>0) th ta c cng thc ca th hm s thay i nh th no?

GV: Gi hc sinh nhn xt tr li ca bn v b sung sai st, ri cho im.Bi mi:

TGHot ng ca GVHot ng ca HSNi dung

GV: Nu cu hi v yu cu hc sinh suy ngh tr li :

Trong mt phng ta Oxy, cho th (G) ca hm s y = f(x); k v l l hai s dng ty . Khi :

a)Nu ta tnh tin th (G) ln trn (theo trc Oy) k n v th c th ca hm s no?

b) Nu ta tnh tin th (G) xung di (theo trc Oy) k n v th c th ca hm s no?

c)Nu ta tnh tin th (G) sang phi (theo trc Ox) l n v th c th ca hm s no?

d)Nu ta tnh tin th (G) sang tri (theo trc Ox) l n v th c th ca hm s no?HS: Nu ta tnh tin th (G) ln trn k n v th ta c th ca hm s y = f(x)+k, cn nus tnh tin xung di k n v th ta c th hm s y =f(x) k.Nu ta tnh tin th (G) sang phi, sang tri theo trc Ox l n v th ta c th ca hm theo th t l: y = f(x-l) v y =f(x+l).

Bng ph:nh l: Trong mt phng ta Oxy, cho th (G) ca hm s y = f(x); k v l l hai s dng ty . Khi .Nu ta tnh tin th (G):

a) Ln trn (theo trc Oy) k n v th c th ca hm s y = f(x) +k.

b) Xung di (theo trc Oy) k n v th c th ca hm s y = f(x) k

c)Sang phi (theo trc Ox) l n v th c th ca hm s y =f(x l).

d) Sang tri (theo trc Ox) l n v th c th ca hm s y = f(x +l).

Bi tp p dng(treo bng ph):Cho hm s y = 4x2-16x +15c th (G) .Nu tnh tin th (G) sang tri 2 n v ta c th ca hm s no?

Nu tip tc tnh tin th (G) ln trn mt n v ta c th ca hm s no?GV: Gi HS nhn xt li gii ca bn v b sung thiu st (nu c).HS: Nu tnh tin th (G) sang tri 2 n v th ta c th ca hm s y =4(x+2)2-16(x+2) +15 = 4x2 1.Tip tc tnh tin th (G) ln trn mt n v ta c th hm s y y =4x2 1+1=4x2.

*Xc nh ng thng:TGHot ng ca GVHot ng ca HSNi dung

GV: Cho 2 ng thng y=ax+b v y =ax+b (a0,a0). Vi iu kin no th hai ng thng cho song song vi nhau?, vung gc vi nhau?

GV: Pht cho cc nhm (nhm l gii cu a v nhm chn gii cu b)v yu cu HS tho lun suy ngh gii trong vng 5 pht sau GV gi HS i din 2 nhm ln bng trnh by li gii.GV: Gi HS cc nhm cn li nhn xt, b sung thiu st (nu c).HS: hai ng thng y=ax+b v y =ax+b song song vi nhau khi v ch khi a=a v b b v vung gc vi nhau khi v ch khi a.a =-1HS nhm 1 trnh by li gii cu a)

th hm s y = ax+b song song vi ng thng y = -2x+1 nu a = -2.Do th i qua im A(2; 2), nn ta c:

2 = -2.2 +bb = 6

Vy hm s cn tm l

Y = -2x + 6.

HS nhm 2 thnh by li gii cu b:

th hm s y = ax+b i qua hai im B(1;1) v C(-1; -5) khi v ch khi:

Vy hm s cn tm l

y=3x-2V d p dng:

Xc nh ng thng y=ax+b, bit th ca n:

a)Song song vi th hm s y = -2x +1 v i qua im A(2;2)

b)i qua hai im B(1;1) v C(-1;-5)

*Xc nh hm s bc hai:TGHot ng ca GVHot ng ca HSNi dung

GV: Cho hm s bc hai y=ax2 +bx+c (a0)

GV Cho HS suy ngh v tr li cc cu hi sau:

nh I c ta nh th no?

th hm s nhn ng thng no lm trc i xng?

Khi a >0 th hm s ng bin, nghch bin trn khong no?Tng t khi a 0 hm s nghch bin trn khong(-;) v ng bin trn khong (; +)HS: V bng bin thin v th

HS: Suy ngh tho lun v trnh by li gii nhm mnh vo bng ph.

HS: i din nhm 3 trnh by li gii.

HS: Nhn xt li gii ca bn v b sung thiu st (nu c).Bng ph vi ni dng:Hm s y =ax2 +bx+c (a0)

Tp xc nh;

nh I;

Trc i xng;

*TH a >0 v a 0 th hm s ng bin v ngc li th nghch bin.

HS: Cc nhm suy ngh tho lun tm li gii trong khong 5 n 7 pht vo bng ph thoe ni dung phn cng.HS: Nhm 1 lp bng bin thin da vo th:

x - 0 +

y + 3 + -1 -1

HS: Nhm 2 trnh by li gii cu b) trn khong

(-;)

Ta c: =(x1+x2)(x12+x22-4)

V x1, x2nn:

Vy hm s nghch bin trn khong (-;)

Trn cc khong cn li gii tng t

HS: Suy ngh so vi bng bin thin.*Phiu HT 2:Ni dung: Hm s y =x4-4x2+3 c th nh hnh v

3

-

O -1a)Da vo th hy lp bng bin thin ca hm s .

b)Tnh t s v xt s bin thin ca hm s trn cc khong ri so snh vi bng bin thin trong cu a).

GV: M0(x0,y0) thuc th hm s y = f(x) khi no?

GV:Cc im trn th hm s y = f(x) c tung l m th honh l nghim ca phng trnh no?

GV: Nu v d p dng v pht phiu hc tp 3, phn cng cng vic cho mi nhm.

GV: Gi HS i din cc nhm cn li nhn xt li gii cu nhm bn v b sung thiu st (nu c).

GV: B sung thiu st nu c v cho im HS theo nhm.HS: im M0(x0,y0) thuc th hm s y = f(x) khi v ch khi x0 thuc tp xc nh ca hm s v y0=f(x0).

HS: Nu cc im trn th hm s y = f(x) c tung l m th honh l nghim ca phng trnh f(x) =m.

HS: Tho lun v tm li gii theo nhm v theo cng vic phn cng.

a)Nhm 3:

iu kin:

Vy tp xc nh l:

b)Nhm 4:im A khng thuc th v xA khng thuc D, im B thuc th, iim C khng thuc, v ta ca im C khng nghim ng

c)Nhm 5: im c tung bng 1 l nghim ca phng trnh =1suy ra: x = 7

Vy im l: M(7;1)*Phiu HT 3:

Ni dung: Cho hm s

.a)Tm tp xc nh ca hm s.

b)Trong cc im A(-2;1), B(1;-1), C(4;2) th im no thuc th hm s?

c)Tm cc im trn th hm s c tung bng 1.

GV: Hng dn v gii cc bi tp 5) 6) 7) 9) 10) v 11, 12 trang 17 trong ti liu t chn nng cao.*Cng c:

*Hng dn hc nh: xem li cc bi tp gii v lm cc bi tp hng dn gii.

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Ch 2PHNG TRNH V H PHNG TRNH (5 Tit)I. MC TIU:

Hc sinh cng c li:

1. V kin thc:

- Nm c kin thc v phng trnh v h phng trnh: Phng trrnh ax +b =0 v phng trnh ax2+bx+c =0, nh l Vi-t v ng dng ca n, h phng trnh bc nht hai n v cch gii.

2. V k nng:

-Gii v bin lun c phng trnh ax +b = 0 v phng trnh ax2+bx+c =0, ng dng ca nh l Vi-t, xt du cc nghim ca phng trnh bc nht v bc hai.

-Gii v bin lun c h phng trnh bc nht hai n, bit cch lp c cc nh thc khi gii h phng trnh v bin lun.

3. V t duy: Rn luyn k nng gii ton, t duy lgic, bit quy l v quen.

4. V thi : Cn thn, chnh xc.

II. CHUN B CA GV V HS:

*i vi HS: Nm vng kin thc v phng trnh v h phng trnh, son bi, n li kin thc hc v lm bi tp trc khi n lp.

*i vi GV: Gio n, bi tp trc nghim, phiu hc tp,

IV. TIN TRNH BI HC V CC HOT NG:

(c chia thnh 5 tit)

Tit 1: n tp kin thc v phng trnh v h phng trnh;Tit 2: Rn luyn k nng gii ton;Tit 3: Rn luyn k nng gii ton v luyn tp;Tit 4: Rn luyn k nng gii ton v luyn tp;

Tit 5: Luyn tp.

---------------(o0o(-----------------

Tit 1: N TP KIN THC V PHNG TRNH V H PHNG TRNH1. n nh lp, chia lp thnh 6 nhm.

2. Kim tra bi c: Kt hp vi iu khin hot ng nhm.Bi mi:

*n tp nhanh kin thc:

TGHot ng ca GVHot ng ca HSNi dung

*Tm tt v b sung kin thc:

A. Phng trnh ax+b=0 v ax2+bx+c=0:

1.Gii v bin lun phng trnh: ax+b=0(1):

GV: Nu cu hi n tp li kin thc c:

-Nu a0 th c nghim khng v nu c th nghim ca phng trnh?

-Nu a =0 th ta phi xt hai trng hp l cc trng hp no?

-Khi b0 th phng trnh nh th no?

-Vy khi b = 0 th phng trnh nh th no?

GV: Treo bng ph tm tt ni dung nu trn.HS: phng trnh c nghim duy nht x=.HS: Trng hp b0 v b=0.

Khi b0 th phng trnh v nghim.

Khi b =0 phng trnh c nghim vi mi x.

HS: Ch theo di ni dung tm tt.Bng ph1:Ni dung:

Gii phng trnh ax+b=0:

*a 0 phng trnh c nghim duy nht x=.

*a =0

b0: phng trnh v nghim

b=0: phng trnh c nghim l x.

B.Phng trnh ax2+bx+c=0(2):

Khi a =0 th phng trnh tr thnh phng trnh ax+b=0 ta bit cch gii v bin lun.

Khi a0 phng trnh (2) l phng trnh bc hai, ta gii bng cch lp , c tnh nh th no?

Phng trnh (2) v nghim, c nghim kp, hai nghim phn bit khi no? Ch ra cng thc nghim.

GV: Hng dn cch gii phng trnh bc 2 bng my tnh b ti.

GV: Nu phng trnh (2) c 2 nghim x1, x2 th ta c phng trnh sau; a(x-x1)(x-x2)=0. V vy ta c ng thc:

ax2+bx+c= a(x-x1)(x-x2)

GV: Treo ghi li ni dung tm tt.HS:=b2-4ac

Phng trnh (2):

+V nghim khi