4 dinh G, H. I, X
4 cgnh GH, HI, IK, KG. Hinh tarn gioc ABC co:
dinh: A, B, C cgnh: As BC, CA
.cgnh
Hinh th' gioc EKST co:
4 dinh .?., ?., *., .?. 4 cqnh: .?., .?., .*., .?.
23 a) Hinh thoy voo .?. mdu gi?
b) Hinh thoy vdo .?. c6 mdy
cgnh?
Em co biet mot logi Iu’dng thu’c mo tén goi co tu'
tarn giâc?
Nu'é'c to ed mot vung
dot trong lua rong 1On cd hinh dong gon nhu‘
mot IO' giac, thu'c‹ng d '
o'c ggi Io tD' gidc Long Xuyén.
e .htho,«. «m conh*
Hinh d6 rndu gt?
Coc thonh pho No Tién, Rach Gio thuoc LTnh Kién Giang. Coc thonh pho Chou Ooc, Long Xuyén thuoc tin h An Gio ng.
X6 tén cdc @t 6trén bum cd dgng cdc hinh khéii dd hoc
Show ru-bfc co
ding khdi Up
phu'ong Khoi ru—bfc v i hoop b0t khzing b| ldn!
Khdi h6p chđ' nhdt. kh6i Up
phu'ong:
• Mai khoi ct:
.?. dinh, .?. conh, .*. mot
•Cdc mdt cua moi
kho Kh6i h§p chs nh@ KhW f§p phmxn@I6
hnh g?
Nhon bi@t dinh, cqnh, m§t cđe khoi h§p chu' nhon Lhiop phu'o'ng.
O0y Id cpnh.
c D0y Id met
Ddy J6 dinh. *
Do I6m ma hinh khoi top phu'ong nhu' hinh
bén, em con: .* due tfnh, .?. vién dot non.
Ba kh6i g6 c6 vdt Id cdc hinh
trén cdt (xem hinh).
Quon sGt céc hinh phbnq v6 hinh kh6i du'd'i d6y.
a) Ké tén cdc hinh phdng. b) Ké tén coc h\nh khoi.
Chon céc hinh noo é' béi dñ ghép lgi thdnh cdc hinh t0' gidc sou?
a) b)
X6p hinh.
Noi md nh gioy (A, B, C, D] Id cdo 6 trong mo trong hinh du'd'i doy?
A B
Hinh thD' nom co boo nhiéu kh6i lop phu'ong?
Hinh Hinh
th0‘ nhot the‘ hoi thr boHinh thu' tu’Hinh
H’inh thai' nom
o can boo nhiéu khoi ) p phu’ong dé xep hinh tu'dng r6o du'é'i d6y?
Vâi cnc pfliñn do k lJoflg phâi lu c nda cñng lo khoi hop chs nh t hay khoi lap phu'ong. bo con nqvo› din t/oc Mong {Iinh Ho Giung) k héo Iéo xéy chi rug thor›h n fling bé’ EU’âr1g rdo rat vñng chs c mo k hong con chot két dinh (voi, via, .. .).
Dém thém 5, theo chiéu mGi
tén, dé biét klm
ph t chi boo nhi5u phut.
5 phut
a yia 5 phut g gif' 20
phdt
8 gid 30 phut hay 8 gid
ru'6'i
8 gio' 35 ph‹Jt 8 gig' JS phdt 8 gif' 55 phut hay 9 gid kém 25 phut hay 9 gid kém 15 phut hay 9 gid kém 5 phut
F1âi ddng hd chi mdy gib? Mou: 9 god 40 phut hay 10 gib kém 20 phdt 5 phut 15 phut IS phut
Xoay kim ding h6 déd6ng h6 chi:
a) 3 gif' ZS phut b) 7 gif' 50 phut c) 11 gid kém 20 phut Chon ddnq h6 phd hg'p vé'i cdch dgc.
MRI ddng h6 chl mgy glb'?
Vdo met bud tgl, hol ddng hd n6o chi c0ng thb'1 glon?
A B C D
Budi sing ngby 20 thdng 11 d' l6'p em.
Ket thuc
a} Chung em bdt ddu lién hoon vbn nghe vdo ldc .?. gio .?. phdt
vo két thdc Iluc .*. gif' kém .?. phut. Deng m6 hinh
b) BOOi Iie'n hoan vdn nghé kéo doi trong .?. phdt. gdr+g hd gym thGm
Tucin cot du' c 1d |o câ. Thu cot du’dc it hdn
Turn 5 Io ct Hoi co hai ban cut du'oc boo nhiéu Io cé?
To1
, o boi gig i,
Bñi gioi
“ So coy to .?. trong du’o'c Io:
So coy .? trong du'oc In:
Noon thdnh tom tcit vd boi gioi.
Tom txt Boi gioi So lo ccr .............?............. Dip so: . ?. 8 cry To 1 trong du c 8 coy, to 2 trong du’ c nhiéu hon to Io 4 coy. Hoi c6 had to tronp d u'o'c boo nhiéu
Anh I 'linh dâ Idt duac 27 vién goch, boc Dñng lot du'o'c nhiéu hdn anh Minh 14 vién gach. H6i cé hai ngu'di da Idt duac bao nhiéu vién goch*
Chon cdch gid1 phu hpp v6'i tdm UL Tdm tdt 1
Cach gidi A - T\m s6 con cd ' b0 thd’ nhot.
' “ OF — Tim I:6ng s6 con cd
d how bé. Be thai nh6t: BE th0' ha:: Tdm tdt 2 10 con ? con Coch gioi B — T\m so con cd ' bé thai hai. 15 con .10 con Be th0‘ hoi:j
@ qué em co nhieu nho nuoi ong mot Trong vu't›'n nhd dnq ngooi co 71 thdng onq, vu'â'n cdo cou Ut ft ho'n vu'd'n cdo onq 16 thung. Hoi trong co hai khu vu'é'n co boo nhiéu thdng ong mcit?
2 cm 3 cm
74 Ti'nh tong do doi
bo m6nh giay. •Biéu thđ‘c
60 - 24; 170 + 65; 5 x 4; 16 : 2; 2 + 2 + 3; 2 x 2 + 3; .. Iâ coc biđ’u thđc.
• Gio tri cua biéu thđ'c
60 - 24 = 36. Gio tri cua biéu thd'c 6D — 24 lo 36. 2 + 2 + 3 = 7. Gio Sri cua biéu th?c 2 + 2 + 3 Io 7.
Tinh gio tr| cdo moi bieu there rdi noi theo mou. Mou! 26 + 35 = 6J
O Tfnh gid tri cdo biéu th0'c.
r 384 + 471 b.' 742 - 42 + 1S9
Moi so la gi6 trot cuo biéu thLr'c noo?
‹:: 187-42 i\• 30: 5 j 70 - 50 f 80 c 2 4 • S 43 + 72 75 — 28 80 + 16 - 22 74 115 47
' Neu trong biéu the chi cé céc phép tinh c§ng, trđ' thi to thu'c hien cdc phép tinh theo th0' tu' th' tréi song phei.
14 — 5 + Z = 9 + 3
» 12
Tfnh gid tri cño bW th0c 10 : 5 x
3.
• N6u trong bleu thđ'c chi cd céc phgp tfnh nhdn, chio thi to thu’c hién cdc phép tinh theo thai' tu’ IN’ tréi sung phdi.
10 : 5 x 3 = 2 x 3
Tfnh gi• *•! •u• biCu thñ'c. a) 82 + 13 — 76 c) 2 = 3 = 5 b) 547 — 264 — 200 d) 16 : 2 : 2 Tfnh g1d t‹! Duo bieu
M6i thdng son d6 ndng 2 kg, mii thung so'n xonh nong 5 kg H6i 4 thung so'n
Cd boo nhiéu
cuon rom nh* Tfnh gif t ! CuO biéu thu'c2 + 5 • 3.
Biéu thee co cdc phép tinh c ng, trñ', nhdn, chio: to thu’c hien coc phép tinh nh6n, chio tru’6'c; rdi thLfc hién coc phép tfnh c@ng, trñ‘ sou.
2 + 5 • 3 = 2 + 15 = 17
Tinh gio tr! cdo bigu th‹?c.
a) 80 — 2 • 7 b) 35 + 12 : 2 c) 45 : 5 — 9
Cou noo ddng, cou noo soi?
a) 70 — 15 + 35 - 90 b/ 50 : 5 x 2 = 20 c) 8 + 2 • 5 = 50 Co tot ma boo nh iéu quñ cñ chu o?
Hoi bi6u I:h0'c dcrt?i doy £hdc nhou 6 dldm ndo7 O6y lb bldu thd'c c6 d6u ngo§c ( ).
Khi tfnh gid tr! cdo bigu th‹?c cé dou ngo§c ( ) thi tru'é'c tién to thus hi4n coc phép tinh trong ngo§c
(2 + 3) x 4 s 5 x 4 = 20
Tfnh gid t•! •u• bidu thGc.
a) 80 - (30 + 25) b) (72 - 67) x B c) 50 : (10 : 2)
Be chuon bi quo tong cho c6c bgn co haort conh kho khbn, Trong xep v6o mii tdi 1 quyen truyén vo 4 quyén vé'. H6i 10 tdi
nhu' vay c6 boo nhiéu quyén truyen vo vé?
No noi vé'i Bi: "To' dâ mua trlg 2 lan, moi lan 3 vi trñ’ng gd vd 1 vi trđ'ng •!*". Biéu th0'c ndo du'di dby giup No tfnh sd vi trñ'nq d6 muo?
VI du: Ouon sdt trén tio sB.
62 65 67 L6m tr6n sd d6n ht›ng chuc
• Sou khi Idmtrt›n sd:d6n hbr›g chyc, hbhg.dan vt,IA ch0’so 0.
Vf d9: 238 250 28A
30 0
Ldm trdn ed dgn hdng chuc.
o) Lom tron coc so 41, 42, 43, d4 den hong chuc thi du'oc so .?.
bl L6m tr6n c6c s6 75, 76, 77, 78, 79 din hdng chuc the du'oc s6 .?. Ldm trt›n so dgn hong bdm.
Lom tron coc so 814, 826, BSS, 847 den hong tram thi diJ’Oc sâ .*.
b) Lom tron cdc so 152, 168, 170, 1B9, 191 dén hdng trdm thi duo'c so .?. L6m trbn cdc s6.
o) Lom mon sd 53 den hdng chuc the dv c sd .?. b› L6m tron so 95 den hong chuc thi du'dc so .?. c) Ldm tr6n sd 620 d6n hong trdm the du'o'c s8.?.
d) Lom tron so 974 den hon so .?.
Tinh t0' v! Sri bié
o) Khoong 200 k thonh ph6 L b) Kho6ng 100
k thdnh ph6 Y6
c) Khoong 50 km n0'o thi dén thonh phd Viet Tr\. d) Kho6ng 10 km nbo th1
ddn thdnh phd Vinh Yén.
Vé'i su' gop su'ccuo rdt nhieu d!• phu'ong vd ngu’é'i don trén cd nu'6'c, hdng trdm c6y edu do dLfo'c xoy dd giup bo con di loi thuon tién.
Ldm tr6n s6 352 ddn hdng tr6m thi
du’oc s6 .?. Mot coy edu duo'c xoy6 tTnh Long An.
Troo doi vé'i ngu'â'i thon de biet quong du'dng IO' nho em dén qué noi, qué ngoo! dai boo nhiéu rdi tom tron so dén hdng chuc.
Bong ho chi
ã MĐt s6 ch0' s6 Lo Id6 thu'fz'ng dix ng:
I: mot V: nom X:
• Cdc s6 Lo F4B t0’ 1 dén 20:
I I T T [ I I\’ \’ V [ VII V fT I IX X
1 2 3 4 5 6 7 8 9 10
XI XII XJIT XIV XV XVI XVII KVILI XIX XP
11 12 13 14 15 16 17 18 19 20
I X. X, XI, XIl, XIII xw' x›', xi'i, xi'ii, x ’iII
xlx, xX
Xooy kim dong hé de dong h6 chi:
Mou: 11 a) 4 g/é'
bl
c) 7 gid d)
Viet coc so sou bong chD' s6 Lo M6.
a) 1, 5, 10 b) 2, 3, 11, 12,
13
c) 4, 6, 14, 16 dt 7, 8, 9, 15, 17, 1B, 19, 20
a} So béy trdm #nh hoi duo'c viet Id:
A. 7 OO2 B. 720 C. 702
b} So 850 du’o'c viét thonh long cdc trdm, chic, don v| lo:
A. 800 + 50 B. 8 + 50 C. 8 + 5 +
0
DQ tinh r6i tlnh.
a) T6ng cda 571 vo 264. b} Hiéu cdo 571 vo 264.
b)2 x (780 - 771)
a) Moi bang hoo co S cdnh hoo. Voy 8 bong hoo co .?. cdnh too. bT C6 12 chiéc d0o nhu' nhau, nhu' v y c6 .?. d6i dña
B6 sinh me ndm b6 25 tudi. Me sinh Tdm n6m me 30 tudi. Nom nay
Tann 9 tuoi. H6i ndm nay bd bao nhiéu tudi? Ho‹sn thi@n cdC cpu sau.
H\nh tarn gidc ABC cd: 3 dinh lo: .?., .?., .?. 3 conh Io: .*., .?., .?. 1km .?. m 1 m .?. dm .?. cm .?. mm 1 dm .?. cm = .?. mm 1 cm = .?. mm 1 mm
b} Sip xép cdc s6 do sou theo th‹r tg' IO' Ié'n dén bé: 2 m, 1 km, 300 cm.
LAM HOP BUT YN vO H0 P OA QUA SP DUNG
Chuon bi
— Mot vâ hop (cho ng han vo hop kem do nh rong) co d on g k hoi hop ch nlña t. — GJciy thu con g, ho din, kéo.
- Thvc/c thong, but c h’i.
2 Thvc hién -Buâc 1
Oot vo h op I 'n mot sou cña té' gioy Thu co ng, ve theo cfc conh cuo khoi hop chñ nhon de dudc 6 hinh ch? nhot (hay hi nh vuonq). - Bcâc 2.
Cnt céc monh giciy hinh chs’ nhot (IJuy h”inh , ’ vuonp) via vé.
— Buñc3.
Don céc mâiJh giciy vu’a co I lén 6 mat c Rio
vo hop.
- Brfo‘c 4:
Cot rnot so h’in lv, don tren coc mot eta h p dc’ try ng I ri.
Hop but cuci em
2. rxc» axé, »xc» cxtn
Tfnh c6 t tfch I rong bo ng nhdn 3. 3 x 1 *. 3 1 = 3 3 x 2 = 6 3 • s -- 9 3 x 5 = 15 3 x 6 - 1g 3 • 7 = 21 3 • 8 = 24 3, 6, 9, 12. 3 • 9 = 27 3 x 10 = 30 3 6 — 9 * - .*. - 18 .*. - 24 — .? 30 A B6 bin coc phép nh6n trong bong nhon 3.
3 • S = 15 3 x . ?. = 21 3 • 7 = 21 Viét phép nh6n. 3 • 1 0 • 3 3 • 0 1 x 3 42
Tfnh nh6m. o) 6:3 ]2:3 9 = 3 • .?. 15-.3. .7. 12 = 3 x .?. 21 = 3 x .?. 12 : 2s .?. 27 3* .7. 18 = 3 x .?. 12 : 3 = 4 15 : 3 = 5 30 : 3 «10 ’ 18 : 3 30 : 3 0 = 3 • .?. 30 = 3 x .?. S6?
a) Chio déu 18 qud didu thdnh 3 phdn, mii phdn c6.?. qud diéu. b) Cd 18 qud di6u chlo thdnh cdc phdn, moi phdn eg 2
qu6. Cd tdt c6.?. phdn.
Viet Nom Id m§t trong cdc node xudt khdu
4 8
3'I h cdc tfch trong Mng nhon1 4 x 1= 4
“ ”