T 1:
T 1:
CC DNG TON
V PHNG PHP GII TON LP 6
Chng 1:
n tp v b tc v s t nhin
Bi 1: Tp hp. Phn t ca tp hp
Dng 1: Vit mt tp hp cho trc
Phng php gii
Dng mt ch ci in hoa v du ngoc nhn, ta c th vit mt tp hp theo hai
cch:
-Lit k cc phn t ca n.
-Ch ra tnh cht c trng cho cc phn t ca n
Dng 2: S dng cc k hiu v
Phng php gii
Nm vng ngha cc k hiu v
K hiu c l phn t ca hoc thuc.
K hiu c l khng phi l phn t ca hoc khng thuc.
Dng 3: Minh ha mt tp hp cho trc bng hnh v
Phng php gii
S dng biu ven. l mt ng cong khp kn, khng t ct, mi phn t ca tp hp
c biu din bi mt im bn trong ng cong .
Bi 2: Tp hp cc s t nhinDng 1: Tm s lin sau, s lin trc ca mt s t
nhin cho trc
Phng php gii
- tm s lin sau ca s t nhin a, ta tnh a+1
- tm s lin trc ca s t nhin a khc 0, ta tnh a-1
Ch : -S 0 khng c s lin trc.
-Hai s t nhin lin tip th hn km nhau 1 n v.
Dng 2: Tm cc s t nhin tha mn iu kin cho trc
Phng php gii
Lit k tt c cc s t nhin tha mn ng thi cc iu kin cho
Dng 3: Biu din trn tia s cc s t nhin tha mn iu kin cho trc
Phng php gii
-Lit k cc s t nhin tha mn ng thi cc iu kin cho
-Biu din cc s va lit k trn tia s
Bi 3: Ghi s t nhin
Dng 1: Ghi cc s t nhin
Phng php gii
-S dng cch tch s t nhin thnh tng lp ghi.
-Ch phn bit: S vi ch s, s chc vi ch s hng chc, s trm vi ch s hng
trm
Dng 2: Vit tt c cc s c n ch s t n ch s cho trc
Phng php gii
Gi s t ba ch s a, b, c khc 0, ta vit cc s c ba ch s nh sau:
Chn a l ch s hng trm ta c: , ;
Chn b l ch s hng trm ta c: , ;
Chn c l ch s hng trm ta c: , .
Vy tt c c 6 s c ba ch s lp c t ba ch s khc 0: a, b v c.
*Ch : Ch s 0 khng th ng hng cao nht ca s c n ch s phi vit.
Dng 3: Tnh s cc s c n ch s cho trc
Phng php gii
tnh s cc ch s c n ch s ta ly s ln nht c n ch s tr i s nh nht c n
ch s ri cng vi 1.
S cc s c n ch s bng: - 1 +1
Dng 4: S dng cng thc m s cc s t nhin
Phng php gii
m cc s t nhin t a n b, hai s lin tip cch nhau d n v. ta dng cng
thc sau:
S cui- s u
Khong cch gia hai s lin tip
+1 ngha l +1
Dng 5: c v vit cc s bng ch s la m
Phng php gii
S dng quy c ghi s La M.
Bi 4: S phn t ca mt tp hp. Tp hp con
Dng 1: Vit mt tp hp bng cch lit k cc phn t theo tnh cht c trng
cho cc phn t ca tp hp y.
Phng php gii
Cn c vo tnh cht c trng cho trc, ta lit k tt c cc phn t tha mn
tnh cht y.
Dng 2: S dng cc k hiu v
Phng php gii
Cn nm vng: K hiu din t quan h gia mt phn t vi mt tp hp; k hiu
din t mt quan h gia hai tp hp.
A M : A l phn t ca M; A M : A l tp hp con ca M.Dng 3: Tm s phn t
ca mt tp hp cho trc
Phng php gii
-Cn c vo cc phn t c lit k hoc cn c vo tnh cht c trng cho cc phn
t ca tp hp cho trc, ta c th tm c s phn t ca tp hp .
- S dng cc cng thc sau:
Tp hp cc s t nhin t a n b c: b a + 1 phn t (1)
Tp hp cc s chn t s chn a n s chn b c: (b a) : 2 + 1 phn t (
2)
Tp hp cc s l t s l m n s l n c: (n-m): 2 + 1 phn t ( 3)
Tp hp cc s t nhin t a n b, hai s k tip cch nhau d n v, c: (b-a):
d +1 phn t
( Cc cng thc (1), (2), (3) l cc trng hp ring ca cng thc (4) )
.
Dng 4: Bi tp v tp rngPhng php gii
Nm vng nh ngha tp hp rng: tp hp khng c phn t no gi l tp hp rng,
k hiu .
Dng 5: Vit tt c cc tp hp con ca tp cho trc
Phng php gii
Gi s tp hp A c n phn t. Ta vit ln lt cc tp hp con:
Khng c phn t no ();
C 1 phn t;
C 2 phn t;
. . .
C n phn t.
Ch : Tp hp rng l tp hp ca mi tp hp: E. Ngi ta chng minh c rng nu
mt hp c n phn t th s tp hp con ca n bng 2n.
Bi 5: Php cng v php nhnDng 1: Thc hnh php cng, php nhn
Phng php gii
-Cng hoc nhn cc s theo hng ngang hoc theo hng dc
-S dng my tnh b ti (i vi nhng bi c php dng )
Dng 2 : p dng cc tnh cht ca php cng v php nhn tnh nhanh
Phng php gii
- Quan st, pht hin cc c im ca cc s hng, cc tha s
- T , xt xem nn p dng tnh cht no (giao hon, kt hp, phn phi) tnh
mt cch nhanh chng.
Dng 3: Tm s cha bit trong mt ng thc
Phng php gii
tm s cha bit trong mt php tnh, ta cn nm vng quan h gia cc s
trong php tnh. Chng hn: s b tr bng hiu cng vi s tr, mt s hng bng
tng ca hai s tr s hng kia
c bit cn ch : vi mi a N ta u c a.0 = 0; a.1=a.
Dng 4: Vit mt s di dng mt tng hoc mt tch
Phng php gii
Cn c theo yu cu ca bi, ta c th vit mt s t nhin cho di dng mt tng
ca hai hay nhiu s hng hoc di dng mt tch ca hai hay nhiu tha s.
Dng 5: Tm ch s cha bit trong php cng, php nhn
Phng php gii
- Tnh ln lt theo ct t phi sang tri. Ch nhng trng hp c nh.
- Lm tnh nhn t phi sang tri, cn c vo nhng hiu bit v tnh cht ca s
t nhin v ca php tnh, suy lun tng bc tm ra nhng s cha bit.
Dng 6: So snh hai tng hoc hai tch m khng tnh c th gi tr ca
chng.
Phng php gii
Nhn xt, pht hin v s dng cc c im ca cc s hng hoc cc tha s trong
tng hoc tch. T da vo cc tnh cht ca php cng v php nhn rt ra kt
lun.
Dng 7: Tm s t nhin c nhiu ch s khi bit iu kin xc nh cc ch s
trong s .
Phng php gii
Da vo iu kin xc nh cc ch s trong s t nhin cn tm tm tng ch s c mt
trong s t nhin .
Bi 6: Php tr v php chia
Dng 1: Thc hnh php tr v php chia
Phng php gii
- C th tr theo hng ngang hoc vit s tr di s b tr sao cho cc ch s
cng hng th thng ct vi nhau ri tr t phi sang tri.
- t php chia v th li kt qu bng php nhn.
- S dng my tnh b ti (i vi nhng bi c php dng).
Dng 2: p dng tnh cht cc php tnh tnh nhanh
Phng php gii
p dng mt s tnh cht sau y:
- Tng ca hai s khng i nu ta thm vo s hng ny v bt i s hng kia cng
mt s n v.
V d: 99 + 48 = (99+1)-( 48-1) = 100+ 47 = 147.
- Hiu ca hai s khng i nu ta thm vo mt s b tr v s tr cng mt s n
v.
V d: 316-97 =(316+3) (97+3) = 319-100= 219
- Tch ca hai s khng i nu ta nhn tha s ny v chia tha s kia cho
cng mt s
V d: 25.12 = (25.4).(12:4) = 100.3 =300
- Thng ca hai s khng i nu ta nhn c s b chia v s chia vi cng mt
s.
V d: 1200: 50 =( 1200.2) : (50.2) =2400:100 =24.
- Chia mt tng cho mt s (a+b) : c = a: c + b:c (trng hp chia
ht).
V d: 276:23 = (230 + 46) : 23 = 230:23 + 46:23 = 10 + 2 =12.
Dng 3: Tm s cha bit trong mt ng thc
Phng php gii
Mun tm mt s hng trong php cng hai s, ta ly tng tr s hng kia;
Mun tm s b tr ta ly hiu cng vi s tr;
Mun tm s tr ta ly s b tr tr i hiu;
Mun tm s b chia ta, ta ly thng nhn vi s chia;
Mun tm s chia, ta ly s b chia chia cho thng.
Dng 4: Bi tp v php chia c d
Phng php gii
S dng nh ngha ca php chia c d v cng thc:
a = b.q + r (0< r < b)
T cng thc trn suy ra : b = (a r) : q; q = (a r) : b; r = a
b.q.
Dng 5: Tm nhng ch s cha bit trong php tr v php chia
Phng php gii
- i vi php tr, tnh ln lt theo ct t phi sang tri, ch nhng trng hp
c nh.
- i vi php chia, t tnh v ln lt thc hin php chia.
Bi 7: Ly tha vi s m t nhin. Nhn hai ly tha cng c s.
Dng 1: Vit gn mt tch bng cch dng ly tha
Phng php gii
p dng cng thc: = an.
Dng 2: Vit mt s di dng mt ly tha vi s m ln hn 1
Phng php gii
p dng cng thc: = an.
Dng 3: Nhn hai ly tha cng c s
Phng php gii
p dng cng thc: am. an = am+n (a, m, n N).
Bi 8: Chia hai ly tha cng c s
Dng 1: Vit kt qu php tnh di dng mt ly tha
Phng php gii
p dng cc cng thc: am . an = am+n; am : an = am-n (a (0, m
(n).
Dng 2: Tnh kt qu php chia hai ly tha bng hai cch
Phng php gii
Cch 1: Tnh s b chia, tnh s chia ri tnh thng.
Cch 2: p dng quy tc chia hai ly tha cng c s ri tnh kt qu.
Dng 3: Tm s m ca mt ly tha trong mt ng thc.Phng php gii
-a v hai lu tha ca cng mt c s.
-S dng tnh cht : vi a ( 0, a ( 1, nu am = an th m = n (a, m, n N
).
Dng 4: Vit mt s t nhin di dng tng cc ly tha ca 10.
Phng php gii
Vit s t nhin cho thnh tng theo tng hng (hng n v, hng chc, hng
trm..). Ch rng 1=100.
V d : 2386 = 2.1000 + 3.100 + 8.10 + 6.1 =2.103 +3.102 + 8.10 +
6.100.
( rng 2.103 l tng hai ly ca 10 v 2.103 = 103 + 103; cng vy i vi
cc s 3.102, 8.10, 6.100 ).
Dng 5: Tm c s ca ly tha
Phng php gii
Dng nh ngha ly tha: = an
Bi 9: Th t thc hin cc php tnhDng 1: Thc hin cc php tnh theo th t
quy nh
Phng php gii
Thc hin theo ng th t quy nh i vi biu thc c du ngoc v biu thc
khng c du ngoc
Dng 2: Tm s cha bit trong ng thc hoc trong mt s
Phng php gii
tm s cha bit trong php tnh, ta cn nm vng quan h gia cc s trong
php tnh.
Ch : Php tnh ngc ca php cng l php tr, php tnh ngc ca php nhn l
php chia.
Dng 3: So snh gi tr hai biu thc i s
Phng php gii
Tnh ring gi tr ca mi biu thc ri so snh hai kt qu tm c.
Bi 10: Tnh cht chia ht ca mt tng
Dng 1: Xt tnh chia ht ca mt tng hoc mt hiu
Phng php gii
p dng tnh cht 1 v tnh cht 2 v s chia ht ca mt tng, mt hiu.
Dng 2: Tm iu kin ca mt s hng tng hoc hiu chia ht cho mt s no
Phng php gii
p dng tnh cht 1 v tnh cht 2 tm iu kin ca s hng cha bit.Dng 3: Xt
tnh chia ht ca mt tch
Phng php gii
p dng tnh cht: Nu trong mt tch cc s t nhin c mt tha s chia ht
cho mt s no th tch cng chia ht cho s .
Bi 11: Du hiu chia ht cho 2 v cho 5
Dng 1: Nhn bit cc s chia ht cho 2 v cho 5
Phng php gii
S dng du hiu chia ht cho 2, cho 5.
S dng tnh cht chia ht ca tng, ca hiu.
Dng 2: Vit cc s chia ht cho 2, cho 5 t cc s hoc cc ch s cho
trc
Phng php gii
Cc s chia ht cho 2 phi c ch s tn cng l 0 hoc 2 hoc 4 hoc 6 hoc
8.
Cc s chia ht cho 5 phi c ch s tn cng l 0 hoc 5.
Cc s chia ht cho 2 v 5 phi c ch s tn cng l 0.
Dng 3: Ton c lin quan n s d trong php chia mt s t nhin cho 2,
cho 5
Phng php gii
* Ch rng:
- S d trong php chia cho 2 ch c th l 0 hoc 1.
- S d trong php chia cho 5 ch c th l 0, hoc1,hoc 2, hoc 3, hoc
4.
Dng 4: Tm tp hp cc s t nhin chia ht cho 2, cho 5 trong mt khong
cho trc.
Phng php gii
Ta lit k tt c cc s chia ht cho 2, cho 5 (cn c vo du hiu chia ht
) trong khong cho.
Bi 12: Du hiu chia ht cho 3, cho 9Dng 1: Nhn bit cc s chia ht
cho 3, cho 9
Phng php gii
S dng du hiu chia ht cho 3, cho 9;
S dng tnh cht chia ht ca tng, ca hiu.
* Ch :
Mt s chia ht cho 9 th cng chia ht cho 3.
Mt s chia ht cho 3 c th khng chia ht cho 9.
Dng 2: Vit cc s chia ht cho 3, cho 9 t cc s hoc cc ch s cho
trc.
Phng php gii
S dng cc du hiu chia ht cho 3, cho 9 (c th c du hiu chia ht cho
2, cho 5)
Dng 3: Ton c lin quan n s d trong php chia mt s t nhin cho 3,
cho 9
Phng php gii
-S dng tnh cht: mt s c tng cc ch s chia ht cho 9 ( cho 3 ) d m
th s chia ht cho 9 (cho 3 ) cng d m
V d : 235 c tng cc ch s bng 2+3+4+5 =14. S 14 chia cho 9 d 5,
chia cho 3 d 2. Do s 2345 chia cho 9 d 5, chia cho 3 d 2.Dng 4: Tm
tp hp cc s t nhin chia ht cho 3, cho 9 trong mt khong cho trc
Phng php gii
-Ta lit k tt c cc s thuc khong cho m c tng cc ch s chia ht cho
3, cho 9 Bi 13: c v bi
Dng 1: Tm v vit tp hp cc c, tp hp cc bi ca mt s cho trc
Phng php gii
- tm c ca mt s, ta chia s ln lt cho 1, 2, 3
- tm bi ca mt s khc 0, ta nhn s ln lt vi 0, 1, 2, 3
Dng 2: Vit tt c cc s l bi hoc c ca mt s cho trc v tha mn iu kin
cho trc
Phng php gii
Tm trong cc s tha mn iu kin cho trc nhng s l bi hoc c ca s
cho.
Dng 3: Bi ton a v vic tm c hoc bi ca mt s cho trc
Phng php gii
Phn tch bi chuyn bi ton v vic tm c hoc bi ca mt s cho trc.
p dng cch tm c hoc bi ca mt s cho trc.
Bi 14: S nguyn t. Hp s. Bng s nguyn t.Dng 1: Nhn bit s nguyn t,
hp s
Phng php gii
Cn c vo nh ngha s nguyn t v hp s.
Cn c vo cc du hiu chia ht.
C th dng bng s nguyn t cui Sgk xc nh mt s (nh hn 1000) l s nguyn
t hay khng.
Dng 2: Vit s nguyn t hoc hp s t nhng s cho trc
Phng php gii
Dng cc du hiu chia ht
Dng bng s nguyn t nh hn 1000.
Dng 3: Chng minh mt s l s nguyn t hay hp s.
Phng php gii
chng minh mt s l s nguyn t, ta chng minh s khng c c no khc 1 v
chnh n.
chng minh mt s l hp s, ta ch ra rng tn ti mt c ca n khc 1 v khc
chnh n. Ni cch khc, ta chng minh s c nhiu hn hai c.
Bi 15 : Phn tch mt s ra tha s nguyn t
Dng 1: Phn tch cc s cho trc ra tha s nguyn t
Phng php gii:
Thng c hai cch phn tch mt s t nhin n (n >1) ra tha s nguyn
t.
Cch 1 (phn tch theo ct dc ): Chia s n cho mt s nguyn t (xt t nh
n ln ), ri chia thng tm c cho mt s nguyn t (cng xt t nh n ln), c
tip tc nh vy cho n khi thng bng 1.
V d: 90 2
45 3
15 3 90 =2.32.5 5 5
1
Cch 2 ( Phn tch theo hng ngang hoc theo s cy ):
90 90 90
2 45 3 30 5 18
9 5 10 3 9 2
3 3 2 5 3 3
90 90
6 15 9 10
2 3 3 5 3 3 2 5
Vit n di dng mt tch cc tha s, mi tha s li vit thnh tch cho n khi
cc tha s u l s nguyn t. V d 90 = 9.10 = 32.2.5.
Tt c cc cch phn tch s 90 ra tha s nguyn t u cho cng mt kt
qu:
90 = 2.32.5.
Dng 2 : ng dng phn tch mt s ra tha s nguyn t tm cc c ca s .
Phng php gii
Phn tch s cho trc ra tha s nguyn t.
Ch rng nu c = a.b th a v b l hai c ca c.
Nh li rng: a = b.q ( a b ( a B(b) ( b U(a) (a,b,q N, b (0)Dng 3:
Bi ton a v vic phn tch mt s ra tha s nguyn t
Phng php gii
Phn tch bi, a v vic tm c ca mt s cho trc bng cch phn tch s ra
tha s nguyn t.
Bi 16: c chung v bi chung
Dng 1: Nhn bit v vit tp hp cc c chung ca hai hay nhiu s
Phng php gii
nhn bit mt s l c chung ca hai s, ta kim tra xem hai s c chia ht
cho s ny hay khng.
vit tp hp cc c chung ca hai hay nhiu s, ta vit tp hp cc c ca mi
s ri tm giao ca cc tp hp .
Dng 2: Bi ton a v vic tm c chung ca hai hay nhiu s
Phng php gii
Phn tch bi ton a v vic tm c chung ca hai hay nhiu s.
Dng 3: Nhn bit v vit tp hp cc bi chung ca hai hay nhiu s
Phng php gii
nhn bit mt s l bi chung ca hai s, ta kim tra xem s ny c chia ht
cho hai s hay khng?
vit tp hp cc bi chung ca hai hay nhiu s, ta vit tp hp cc bi ca
mi s ri tm giao ca cc tp hp .
Dng 4: Tm giao ca hai tp hp cho trc
Phng php gii
Chn ra nhng phn t chung ca hai tp hp A v B. chnh l cc phn t ca A
B.
Bi 17: c chung ln nht
Dng 1: Tm c chung ln nht ca cc s cho trc
Phng php gii
Thc hin quy tc ba bc tm CLN ca hai hay nhiu s.
Dng 2: Bi ton a v vic tm CLN ca hai hay nhiu s
Phng php gii
Phn tch bi, suy lun a v vic tm UCLN ca hai hay nhiu s
Dng 3: Tm cc c chung ca hai hay nhiu s tha mn iu kin cho trc
Phng php gii
Tm CLN ca hai hay nhiu s cho trc;
Tm cc c ca CLN ny;
Chn trong s cc c tha mn iu kin cho.
Bi 18: Bi chung nh nht Dng 1: Tm bi chung nh nht ca cc s cho trc
Phng php gii
Thc hin quy tc ba bc tm BCNN ca hai hay nhiu s.
C th nhm BCNN ca hai hay nhiu s bng cch nhn s ln nht ln lt vi
1,2, 3, cho n khi c kt qu l mt s chia ht cho cc s cn li.
Dng 2: Bi ton a v vic tm BCNN ca hai hay nhiu s.
Phng php gii
Phn tch bi, suy lun a v vic tm BCNN ca hai hay nhiu s.
Dng 3: Bi ton a v vic tm bi chung ca hai hay nhiu s tha mn iu
kin cho trc Phng php gii
Phn tch bi, suy lun a v vic tm bi chung ca hai hay nhiu s cho
trc
Tm BCNN ca cc s ;
Tm cc bi ca cc BCNN ny;
Chn trong s cc bi tha mn iu kin cho.
CHNG II: S NGUYN
Bi 1: Lm quen vi s nguyn m
Dng 1: Hiu ngha ca vic s dng cc s mang du (
Phng php gii
Nm vng quy c v ngha ca cc s mang du (, v d dng biu th nhit di
0oC, su di mc nc bin
Dng 2: Ghi cc im biu din s nguyn trn trc s
Phng php gii
Trn trc s, cc im biu din s nguyn m nm bn tri im gc; cc im biu
din s t nhin khc 0 nm bn phi im gc.
Bi 2: Tp hp cc s nguyn
Dng 1: c v hiu ngha cc k hiu , N, Z
Phng php gii
Cn c vo ngha cc k hiu, pht biu bng li v xc nh tnh ng sai ca vic
s dng k hiu.
Dng 2: Hiu ngha ca vic s dng cc s mang du + v cc s mang du ( biu
th cc i s c hai hng ngc nhau. Phng php gii
Trc ht cn nm vng quy c v ngha ca cc s mang du + v cc s mang du (
(quy c ny thng c nu trong bi )
V d: Vit +50C ch nhit 5o trn 0oC, vit -5oC ch nhit 5o di
0oC.
Trn c s quy c , pht biu bng li hoc biu din bng im trn trc s.
Dng 3: Tm s i ca cc s cho trc
Phng php gii
Ch rng hai s i nhau ch khc nhau v du.
S i ca s 0 l 0
Bi 3: Th t trong tp hp cc s nguyn
Dng 1: So snh cc s nguyn
Phng php gii
Cch 1:
Biu din cc s nguyn cn so snh trn trc s;
Gi tr cc s nguyn tng dn t tri sang phi.
Cch 2: Cn c vo cc nhn xt sau:
S nguyn dng ln hn 0;
S nguyn m nh hn 0;
S nguyn dng ln hn s nguyn m;
Trong hai s nguyn dng, s no c gi tr tuyt i ln hn th s y ln
hn;
Trong hai s nguyn m, s no c gi tr tuyt i nh hn th s y ln hn.
Dng 2: Tm cc s nguyn thuc mt khong cho trc
Phng php gii
V trc s v th hin khong cho trc trn trc s;
Tm trn trc s cc s nguyn thuc khong cho.
Dng 3: Cng c khi nim gi tr tuyt i ca mt s nguyn
Phng php gii
Vic gii dng ton ny cn da trn cc kin thc sau v gi tr tuyt i ca mt
s nguyn:
Gi tr tuyt i ca mt s t nhin l chnh n;
Gi tr tuyt i ca mt s nguyn m l s i ca n;
Gi tr tuyt i ca mt s nguyn l mt s t nhin;
Hai s nguyn i nhau c cng mt gi tr tuyt i.
Dng 4: Cng c li v tp hp N cc s t nhin v tp hp Z cc s nguyn
Phng php gii
Cn nm vng : N = { 0; 1; 2; 3; 4; .};
Z = {-3; -2; -1; 0; 1; 2; 3; .}.
Dng 5: Bi tp v s lin trc, s lin sau ca mt s nguyn
Phng php gii
Cn nm vng: s nguyn b gi l s lin sau ca s nguyn a nu a < b v
khng c s nguyn no nm gia a, b; khi , ta cng ni a l s lin trc ca
b
Bi 4: Cng hai s nguyn cng du
Dng 1: Cng hai s nguyn cng du
Phng php gii
p dng quy tc cng hai s nguyn cng du.
Dng 2: Bi ton a v php cng hai s nguyn cng du
Phng php gii
Phn tch bi a v php cng hai s nguyn cng du.
Dng 3: in du >, < thch hp vo vung
Phng php gii
p dng quy tc cng hai s nguyn cng du ri tin hnh so snh hao s
nguyn
Bi 5: Cng hai s nguyn khc du
Dng 1: Cng hai s nguyn
Phng php gii
p dng quy tc cng hai s nguyn cng du v quy tc cng hai s nguyn khc
du.
Dng 2: Bi ton a v php cng hai s nguyn
Phng php gii
Cn c vo yu cu ca bi, thc hin php cng hai s nguyn cho trc
Dng 3: in s thch hp vo trng
Phng php gii
Cn c vo quan h gia cc s hng trong mt tng v quy tc cng hai s
nguyn ( cng du, khc du ), ta c th tm c s thch hp
Bi 6 : Tnh cht ca php cng cc s nguyn
Dng 1: Tnh tng cc nhiu s nguyn cho trc
Phng php gii
Ty c im tng bi, ta c th gii theo cc cch sau : - p dng tnh cht
giao hon v kt hp ca php cng - Cng dn hai s mt
- Cng cc s dng vi nhau, cng cc s m vi nhau, cui cng cng hai kt
qu trnDng 2 : Tnh tng tt c cc s nguyn thuc mt khong cho trc
Phng php gii
- Lit k tt ccc s nguyn trong khong cho trc
- Tnh tng tt c cc s nguyn , ch nhm tng cp s i nhau
Dng 3 : Bi ton a v php cng cc s nguyn
Phng php gii
Cn c vo ni dung ca bi, phn tch a bi ton v vic cng cc s nguyn
Dng 4 : S dng my tnh b ti cng cc s nguyn
Phng php gii
+
-
Khi dng my tnh b ti cng cc s nguyn, cn ch s dng ng nt
.(xem hnh dn s dng trong SGK trang 80 )
Bi 7: Php tr hai s nguynDng 1: Tr hai s nguyn
Phng php gii
p dng cng thc: a b = a + (-b)
Dng 2 : Thc hin dy cc php tnh cng, tr cc s nguyn
Phng php gii
Thay php tr bng php cng vi s i ri p dng quy tc cng cc s nguyn
Dng 3 : Tm mt trong hai s hng khi bit tng hoc hiu v s hng kia
Phng php gii
S dng mi qua h gia cc s hng vi tng hoc hiu
- Mt s hng bng tng tr s hng kia ;
- S b tr bng hiu cng s tr ;
- S tr bng s b tr tr hiu ;
i vi nhng bi n gin c th nhm kt qu ri th li.
Dng 4 : Tm s i ca mt s cho trc
Phng php gii
p dng : s i ca a l a. Ch : -(-a) = a
Dng 5 : vui lin quan n php tr s nguyn
Phng php gii
Cn c vo yu cu ca bi suy lun dn n php tr hai s nguyn
Bi 8 : Quy tc du ngoc
Dng 1 : Tnh cc tng i s
Phng php gii
Thay i v tr s hng v b ngoc hoc du ngoc mt cch thch hp ri
tnh.
Dng 2 : p dng quy tc du ngoc n gin biu thc
Phng php gii
B du ngoc ri thc hin php tnh.
Bi 9 :Quy tc chuyn v
Dng 1 : Tm s cha bit trong mt ng thc
Phng php gii
p dng tnh cht ca ng thc, quy tc du ngoc v quy tc chuyn v ri thc
hin php tnh vi cc s bit.
Dng 2: Tm s cha bit trong mt ng thc c cha du gi tr tuyt i
Phng php gii
Cn nm vng kha nim gi tr tuyt i ca mt s nguyn a. l khong cch t im
a n im 0 trn trc s (tnh theo n v di lp trc s).
Gi tr tuyt i ca s 0 l s 0.
Gi tr tuyt i ca mt s nguyn dng l chnh n;
Gi tr tuyt i ca mt s nguyn m l s i ca n ( v l mt s nguyn
dng).
Hai s i nhau c gi tr tuyt i bng nhau.
T suy ra = a (aN ) th x = a hoc x = -a.
Dng 3: Tnh cc tng i s
Phng php gii
Thay i v tr s hng, p dng quy tc du ngoc mt cch thch hp ri lm php
tnh.Bi 10: Nhn hai s nguyn khc duDng 1 : Nhn hai s nguyn khc du
Phng php gii
p dng quy tc nhn hai s nguyn khc du.
Dng 2: Bi ton a v thc hin php nhn hai s nguyn khc du.
Phng php gii
Cn c vo bi, suy lun dn n vic thc hin php nhn hai s nguyn khc
du.
Dng 3: Tm cc s nguyn x, y sao cho x.y = a (a Z , a 0, b>0
- Mau b cho biet so phan bang nhau ma hnh c chia ra ;
- T a cho biet so phan bang nhau a lay.
Dang 2: Viet cac phan so
Phng phap giai :
- a phan b , a:b c viet thanh .- Chu y rang trong cach viet , b
phai khac 0.
Dang 3: Tnh gia tr cua phan so
Phng phap giai :
e tnh gia tr cua mot phan so, ta tnh thng cua phep chia t cho
mau. Khi chia so nguyen a cho so nguyen b (b( 0) ta chia cho roi at
dau nh trong quy tac nhan hai so nguyen.
Dang 4: Bieu th cac so o theo n v nay di dang phan so theo n v
khac.
Phng phap giai :
e giai dang toan nay, can nam vng bang n v o lng : o o dai, o
khoi lng, o dien tch, o thi gian.
Chang han : 1dm = m ; 1g = kg ; 1cm = m ;
1dm = m ; 1s = h ;
Dang 5: Tm iu kin phn s tn ti iu kin phn s c gi tr l s nguyn
Phng phap giai :
- Phan so ton tai khi t va mau la cac so nguyen va mau khac
0.
- Phan so co gia tr la so nguyen khi mau la c cua t.
Bai 2. PHAN SO BANG NHAUDang 1: Nhan biet cac cap phan so bang
nhau, khong bang nhau
Phng phap giai :
- Neu a.d = b.c th = ;
- Neu a.d ( b.c thi (;
Dang 2: Tm so cha biet trong ang thc cua hai phan so
Phng phap giai :
= nen a.d = b.c (nh ngha hai phan so bang nhau).
Suy ra : a = , d = , b = , c = .
Dang 3: Lap cac cap phan so bang nhau t mot ang thc cho trc
Phng phap giai :
T nh ngha hai phan so bang nhau ta co :
a.d = b.c = ;
a.d = c.b = ;
d.a = b.c = ;
d.a = c.b = ;
Bai 3. TNH CHAT C BAN CUA PHAN SO
Dang 1: Ap dung tnh chat c ban cua phan so e viet cac phan so
bang nhau
Phng phap giai
Ap dung tnh chat : = (m Z, m( 0) ;
= (n C(a,b)).
Dang 2: Tm so cha biet trong ang thc cua hai phan so
Phng phap giai :
Ap dung tnh chat c ban cua phan so e bien oi hai phan so a cho
thanh hai phan so bang chung nhng co t (hoac mau) nh nhau. Khi o,
mau (hoac mau) cua chung phai bang nhau, t o tm c so cha biet .
Dang 3: Giai thch l do bang nhau cua cac phan so
Phng phap giai :
e giai thch l do bang nhau cua cac phan so, ta co the :
- Ap dung tnh chat c ban cua cac phan so e bien phan so nay
thanh phan so kia hoac bien ca hai phan so thanh mot phan so th
ba.
- S dung nh ngha phan so bang nhau (xet tch cua t phan so nay vi
mau cua phan so kia).
Bai 4: RUT GON PHAN SO
Dang 1: Rut gon phan so. Rut gon bieu thc dang phan so
Phng phap giai :
- Chia ca t va mau cua phan so cho CLN cua va e rut gon phan so
toi gian.- Trng hp bieu thc co dang phan so, ta can lam xuat hien
cac tha so chung cua t va mau roi rut gon cac tha so chung o.
Dang 2: Cung co khai niem phan so co ket hp rut gon phan so
Phng phap giai :
Can c vao y ngha cua mau va t cua phan so (trng hp mau va t la
cac so nguyen dng) e giai, chu y rut gon khi phan so cha toi
gian.
Dang 3. Cung co khai niem hai phan so bang nhau
Phng phap giai :
- S dung nh ngha hai phan so bang nhau.
- S dung tnh chat c ban cua phan so; quy tac rut gon phan
so.
Dang 4: Tm phan so toi gian trong cac phan so cho trc
Phng phap giai :
e tm phan so toi gian trong cac phan so cho trc, ta tm CLN cua
cac gia tr tuyet oi cua t va mau oi vi tng phan so. Phan so nao co
CLN nay la 1 th o la phan so toi gian.
V du : Phan so toi gian v CLN ( , ) = CLN (5,7) =1.
Dang 5: Viet dang tong quat cua tat ca cac phan so bang mot phan
so cho trc
Phng phap giai :
Ta thc hien hai bc :
- Rut gon phan so a cho en toi gian, chang han c phan so toi
gian ;
- Dang tong quat cua cac phan so phai tm la (k , k ( 0).
Dang 6: Chng minh mot phan so la toi gian
Phng phap giai :
e chng minh mot phan so la toi gian, ta chng minh CLN cua t va
mau cua no bang 1 (trng hp t va mau la cac so nguyen dng; neu la so
ngueyen am th ta xet so oi cua no).
Bai 5. QUY ONG MAU NHIEU PHAN SO
Dang 1: Quy ong mau cac phan so cho trc
Phng phap giai :
Ap dung quy tac quy ong mau nhieu phan so vi mu dng .
* Chu y : Trc khi quy ong can viet cac phan so di dang phan so
vi mau dng. Nen rut gon cac phan so trc khi thc hien quy tac .
Dang 2: Bai toan a ve viec quy ong mau nhieu phan so
Phng phap giai :
C an c vao ac iem va yeu cau cua e bai e a bai toan ve viec quy
ong mau cac phan so .
Bai 6. SO SANH PHAN SO
Dng 1: So sanh cac phan so cung mau
Phng phap giai :
- Viet phan so co mau am thanh phan so bang no va co mau
dng.
-So sanh cac t cua cac phan so co cung mau dng, phan so nao co t
ln hn th ln hn .
Dang 2: So sanh cac phan so khong cung mau
Phng phap giai :
- Viet phan so co mau am thanh phan so bang no va co mau dng
-Quy ong mau cac phan so co cung mau dng
-So sanh t cua cac phan so a quy ong
Bai 7: PHEP CONG PHAN SO
Dang 1: Cong hai phan so
Phng phap giai:-Ap dung quy tac cong hai phan so cung mau ,quy
tac cong hai phan so khong cung mau .
-Nen rut gon phan so (neu co phan cha toi gian ) trc khi cong
.chu y rut gon ket qua (neu co the ).
Dang 2: ien dau thch hp( ,= ) vao o vuong
Phng phap giai:
Thc hien phep cong phan so roi tien hanh so sanh.
Dang 3: Tm so cha biet trong mot ang thc co cha phep phep cong
phan so.
Phng phap giai :Thc hien phep cong phan so roi suy ra so phai
tm.
Dang 4: So sanh phan so bang cach s dung phep cong phan so thch
hp .
Phng phap giai :
Trong mot so trng hp e so sanh hai phan so ,ta co the cong chung
vi hai phan so thch hp co cung t. So sanh hai phan nay se giup ta
so sanh c hai phan so a cho .
Khi so sanh hai phan so cung t can chu y :
-Trong hai phan so co cung t dn , phan so nao co mau ln hn th
phan so nao nho hn ;
-Trong hai phan so co cung t am, phan so nao co t ln hn th ln
hn
Bai 8. TNH CHAT C BAN CUA
PHEP CONG PHAN SO
Dang1 : Ap dung cac tnh chat cua phep cong e tnh nhanh tong cua
nhieu phan so
Phng phap giai:
e tnh mot cach nhanh chong cac cho trc, ta thng can c vao ac iem
cua cac so hang e ap dung cac tnh chat giao hoan va ket hp cua phep
cong mot cach hp l.
Dang 2: Cong nhieu phan so
Phng phap giai:
Nh tnh chat ket hp ,ta co the m rong quy tac cong hai phan so e
cong t ba phan so tr len.
Dang 3: Ren luyen k nang cong hai phan so
Phng phap giai :
Cac bai tap dang nay c trnh bay di nhieu hnh thc khac nhau song
eu oi hoi phai k nang cong phan so thanh thao ,co khi con nham e d
oan so hang con thieu trong phep cong ,hoac phap hien cho sai khi
lam tnh .
Bai 9. PHEP TR PHAN SO
Dang 1: Tm so oi cua mot so cho trc .
Phng phap giai :
e tm so oi cua mot so khac 0 ,ta ch can oi dau cua no .
chu y: so oi cua so 0 la 0.
Dang 2: Tr mot phan so cho mot phan so
Phng phap giai :
Ap dung quy tac thc hien phep tr phan so :.
Dang 3: Tm so hang cha biet trong mot tong ,mot hieu
Phng phap giai : Chu y quan he gia cac so hang trong mot tong
,mot hieu
- Mot so hang bang tong tr i so hang kia ;
- So b tr bang hieu cong vi so tr ;
- So tr bang so b tr tr i hieu .
Dang 4: Bai toan dan en phep cong phep tr phan so
Phng phap giai :
Can c vao e bai ,lap cac phep cong, phep tr phan so thch hp
.
Dang 5: Thc hien mot day tnh cong va tnh tr phan so
Phng phap giai : Thc hien cac bc sau :
-Viet phan so co mau am thanh phan so bang no va co mau dng
;
- Thay phep tr bang phep cong vi so oi ;
- Quy ong mau cac phan so roi thc hien cong cac t ;
- Rut gon ket qu.
Tuy theo ac im cua cac phan so, co the ap dung cac tnh chat cua
phep cong phan so e viec tnh toan c n gian va thuan li.
Bai 10. PHEP NHAN PHAN SO
Dang 1: Thc hien phep nhan phan so
Phng phap giai :
Ap dung quy tac nhan phan so .nen rut gon (neu co the ) trc va
sau khi lam tnh nhan .
Dang 2: Viet mot phan so di dang tch cua hai phan so thoa man
ieu kien cho trc
Phng phap giai :
-Viet cac so nguyen t va mau di dang tch cua hai so nguyen ;
- Lap cac phan so co t va mau chon trong cac so nguyen o sao cho
chung thoa man ieu kien cho trc .
Dang 3: Tm so cha biet trong mot ang thc co cha phep nhan phan
so .
-Thc hien phep nhan so
-Van dung quan he gia cac so hang vi tong hoac hieu trong phep
cong, phep tr .
Dang 4: So sanh gia tr hai bieu thc
Phng phap giai:
Thc hien phep tnh ( cong ,tr ,nhan phan so )e tnh gia tr hai
bieu thc roi so sanh hai ket qua thu c .
Bai 11. TNH CHAT C BAN CUA PHEP NHAN PHAN SO
Dang 1: Thc hien phep nhan phan so
Phng phap giai :- Ap dung quy tac phep nhan phan so ;
- Van dung tnh chat c ban cua phep nhan phan so khi co the .
* chu y:
Dang 2: Tnh gia tr bieu thc
Phng phap giai : - Chu y thc hien cac phep tnh :
a) oi vi bieu thc khong co dau ngoac ;
Luy tha nhancong va tr .
b) oi vi bieu thc co dau ngoac :
( ) [ ] { }.
-Ap dung cac tnh chat c ban cua phan so khi co the .
Dang 3: Bai toan dan en phep nhan phan so
Phng phap giai :
Can c vao e bai, lap phep nhan phan so thch hp .
Bai 12. PHEP CHIA PHAN SO
Dang 1: Tm so nghch ao cua mot so cho trc
Phng phap giai:
- Viet so cho trc di dang ( a,b Z, a0,b0 ).
- So nghch ao cua la .
- So 0 khong co so nghch ao .
-So nghch ao cua so nguyen a (a0) la .
Dang 2: Thc hien phep chia phan so
Phng phap giai:
-Ap dung quy tac chia mot phan so hay mot so nguyen cho mot phan
so
-Khi chia mot phan so cho mot so nguyen ( khac 0), ta gi nguyen
t so cua phan so va nhan mau vi so nguyen .
Dang 3: Viet mot phan so di dang thng cua hai phan so thoa man
ien kien cho trc
Phng phap giai:
- Viet cac so nguyen t va mau di dang tch cua hai so nguyen.
- Lap cac phan so co t va mau chon trong cac so nguyen o sao cho
chung thoa man ieu kien cho trc ;
- Chuyen phep nhan phan so thanh phep chia cho so nghch ao.
Dang 4: Tm so cha biet trong mot tch , mot thng
Phng phap giai :
Can xac nh quan he gia cac so trong phep nhan, phep chia :- Muon
tm mot trong hai tha so, ta lay tch chia cho tha so kia;
- Muon tm so b chia, ta lay thng nhan vi so chia ;
- Muon tm so chia, ta lay so b chia chia cho thng .
Dang 5: Bai toan dan en phep chia phan so
Phng phap giai :
C an c vao e bai, ta lap phep chia phan so, t o hoan thanh li
giai cua bai toan.
Dang 6: Tnh gia tr cua bieu thc
Phng phap giai :
Can chu y th t thc hien cac phep tnh : Luy tha roi en nhan,
chia, cong, tr. Neu co dau ngoac, ta thng lam phep tnh trong ngoac
trc .
Khi chia mot so cho mot tch, ta co the chia so o cho tha so th
nhat roi lay ket qua o chia tiep cho tha so th hai : a: ( b.c) =
(a:b) :c
Bai 13. HON SO. SO THAP PHAN. PHAN TRAM
Dang 1: Viet phan so di dang hon so va ngc lai
Phng phap giai :
Ap dung quy tac viet phan so di dang hon so va quy tac viet hon
so di dang phan so .
Dang 2: Viet cac so a cho di dang phan so thap phan. So thap
phan, phan tram va ngc lai.
Phng phap giai :
Khi viet can lu y : So ch so cua phan thap phan phai ung bang so
0 mau cua phan so thap phan.
Dang 3: Cong, tr hon so
Phng phap giai :
-Khi cong hai hon so ta co the viet chung di dang phan so roi
thc hien phep cong phan so. Ta co the cong phan nguyen vi nhau,
cong phan phan so vi nhau (khi hai hon so eu dng).
V du: 2+3= (2+3) + () =5+=5
- Khi tr hai hn s, ta c th vit chng di dng phn s ri thc hin php
tr phn s. Ta cng c th ly phn nguyn ca s b tr tr phn nguyn ca s tr,
phn phn s ca s b tr tr phn phn s ca s tr, ri cng kt qu vi nhau (khi
hai hn s u dng, s b tr ln hn hoc bng s tr)
V d : 3 - 2= (3-2) +(-) = 1+= 1
-Khi hai hn s u dng, s b tr ln hn hoc bng s tr nhng phn phn s ca
s b tr nh hn phn phn s ca s tr, ta phi ri mt n v phn nguynca s b tr
thm vo phn phn s, sau tip tc tr nh trn
V d : 8-3= 8-3= 7-3= 4
Dng 4 : Nhn, chia hn s
Phng php gii
-Thc hin php cng hoc php tr hn s bng cch vit hn s di dng phn s
ri lm php cng hoc php chia phn s.
-Khi nhn hoc chia mt hn s vi mt s nguyn, ta c th vit hn s di dng
mt tng ca mt s nguyn v mt phn s.
V d : 2.2 = (2+).2 = 2.2 +.2 = 4+= 4
6: 2 = (6+) : 2= 6: 2+:2 = 3+ 1/5 = 3
Dng 5: Tnh gi tr ca biu thc s
Phng php gii
tnh gi tr ca biu thc s ta cn ch :
- Th t thc hin cc php tnh.
- Cn c vo c im ca cc biu thc c th p dng tnh cht cc php tnh v quy
tc du ngoc.
Dng 6: Cc php tnh v s thp phn
Phng php gii
- S thp phn c th vit di dng phn s v ngc phn s cng vit dc di dng
s thp phn.
- Cc php tnh v soos thp phn cng c cc tnh cht nh php tnh v phn
s.
Bi 14: Tm gi tr phn s ca mt s cho trc
Dng 1: Tm gi tr phn s ca mt s cho trc
Phng php gii
tm gi tr phn s ca mt s cho trc, ta nhn s cho trc vi phn s
Phn s c th c vit di dng hn s, s thp phn, s phn trm
ca s b l : b. ( m, n N, n ( 0);
Dng 2: Bi ton dn n tm gi tr phn s ca mt s cho trc
Phng php gii
Cn c vo ni dung c th ca tng bi, ta phi tm gi tr phn s ca mt s
cho trc trong bi, t hon chnh li gii ca bi ton.
Bi 15: Tm mt s bit gi tr mt phn s ca n.
Dng 1: Tm mt s bit gi tr mt phn s ca n.
Phng php gii
Mun tm mt s bit gi tr mt phn s ca n, ta chia gi tr ny cho phn
s
ca s x bng a, th x = a : (m, n N* ).
Dng 2: Bi ton dn n tm mt s bit gi tr mt phn s ca n
Phng php gii
Cn c vo bi, ta chuyn bi ton v tm mt s bit gi tr mt phn s ca n, t
tm c li gii bi ton cho.
Dng 4: Tm s cha bit trong mt tng, mt hiu.
Phng php gii Cn c vo quan h gia s cha bit v cc s bit trong php
cng, php tr tm s cha bit.
Bi 16 : Tm t s ca hai s
Dng 1: Cc bi tp c lin quan n t s ca hai s
Phng php gii
tm t s ca hai s a v b, ta tnh thng a:b
Nu a v b l cc s o th chng phi c o bng cng mt dn v.
Dng 2: Cc bi tp lin quan n t s phn trm
Phng php gii
C ba bi ton c bn v t s phn trm:
1. Tm p% ca s a : x = . a = .
2. Tm mt s bit p% ca n l a: x = a: =
3. Tm t s phn trm ca hai s a v b: = %
Dng 3: Cc bi tp c lin quan n t l xch
Phng php gii
C ba bi ton c bn v t l xch.
Nu gi t l xch l T, khong cch gia hai im trn bn v l a, khong cch
gia hai im tng ng trn thc t l b th ta c bi ton c bn sau:
1. Tm T bit a v b: T = .
2. Tm a bit T v b : a = b.T.
3. Tm b bit T v a : b = .
* Ch : a v b phi cng n v o.
Bi 17: Biu phn trm
Dng 1: Dng biu phn trm theo cc s liu cho trc
Phng php gii
Cn c vo cc s liu phn trm cho, dng biu phn trm theo yu cu ca
bi.
Dng 2: c biu cho trc
Phng php gii
Trn c s hiu ngha ca cc biu , cn c vo biu cho m rt ra nhng thng
tin cha ng trong biu .
Dng 3: Tnh t s phn trm ca cc s cho trc
Phng php gii
p dng quy tc tm t s phn trm ca hai s.
i vi nhng s ln c th dng my tnh b ti.
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