[r]
(1)so cdo DUC vA EAo rAo AN GIANG
on cniNn unrc s6 bao danh
oE ftil:
BAI 1: (5di6m)
Kh6ng phdi ldm tinh,
a) 2l l2x2
r c)0,4 1 10,4x0,4
ri rHr HQC sINH GI6I TIEU IIQC
Ngdy thi : 181 .312012
MOn thi : TOAN
Thoi gian : 90 Phft
-ooOoo-em hdy di6n ddu ") , ( , =" vAo tr6ng:
11 -Xaata
1-b);l I
d)123,45x6,789 l f oza,sx12,345
cm2
e) 38 + 2x0,3 : 0,4 lZ: 0,4 x 0,3 + 38
BAI 2: (5 di6m) a) Tinh:
ii iA;'+ 2kg , kg; 13m + 4cm = '.""cm; 14m2 + 5cm2 =
b) Cho a, b, c lir c6c s6 tg nhi6n lcvn hon 0, ta c6: a - b = c'
H6y so sdnh c6c s5: a vd b; a vd c; b vd c'
BAI 3: (5 il6m)
MOtcfra nang c6 6 bao gqo M6i bao cin nang lAn luqt ld 15k9, 19k9,.20k9'
21kg,27kgvd 28kg Cua hAng.dd b6n,5 bao g4o cho 2 ngucri kh6ch' 56
ki-10-g.r-gao n-guoi tnf"ntrAt ru" glSp d6i s5 ki-lo-gam'nguoi thu hai mua H6i: a) Cdn lai bao gqo n?ro.chua bAn?
bi Nguoi kh6ch thrjr nhAt mua nh[vng bao gqo ndo? Ngucri kh6ch thu hai mua nh0ng bao gqo ndo?
gAt +: U di6m)
Hinh ch0 nhit ABCD c6 chiAu dai AB ld 4cm, chidu rQng BC ld 3cm.
a) HaY v6 hai hinh ch0 nhit co ctng
di€n t(ch vcvi hinh ch0 nhit ABCD nhung co chi6u diri, chi6u rOng kh6c voi hinh ch0 nh?t ABCD.
b) Tr€n cqnh AB ldY di6m P, trOn canh
CD l6v di6m Q cho AP = CQ' GPi M
la trung di6m c0a BC' Tinh dien tich PBCQ, diQn tich PMQ.
(2)so crAo DUC vA EAo rao ri rm HQC SINH GIOI npu HQC Ngdy thi : I8l3l20l2
-g:1*-o-
-ooooo-Hufng dfin ch6m mdn TOAN
BAI 1: (5 di6m)
Kh6ng phii ldm tinh, em hdy di6n d5u "> , < Htdng gidi:
a) 2l <12x2
c; o,+ | > lo ,4 xo,4
e) 38 + x 0,3 : o,+ [-jl 2:0,4x 0,3 + 38 Bieu diem:
DiSn d0ng m6i6 dugc 1 di6m
BAf 2: (5 di6m) a) Tinh:
Di6n sai m6io ul tr} 1 di6m cia Bii 1
14m2 + 5cm2 =
b-c.
=" vdo tr6no
l- l
b) ,?, :l rl : x :, j j
d) 't-l123,45 x 6,7891 - | 678,9 x 12,345
.cmt
11 tAn + 2kg - rkg; 13m + 4cm cm; b) Cho a, b, c ld c6c s6 tu nhi6n ldn hon 0, ta c6: a
-Hiy so s6nh c6c s6: a vA b; a vdr c; b vir c Htdng gidi:
a) Tinh:
tt t6n + 2kg - 11002kg; 13m + 4cm = 1304cm; 14m2 + 5cm2 = 140005cm2
b) a-b= cn6n a > b, a > ccdn bvdcchuax6cdinh duEcs6 ndo l6n hon s6 ndo 6reu otem:
Clru a (3d): Tinh dring m6i pnep tinh duqc 1 ,0 d
CAub GA: N€u dugc a>b (0,5 tl); n6u duqc a>c (0,5 d); n6u duEc b vd c chua x6c eflnh (1d)
BAI 3: (5 di6m)
MOt cira hdng c6 6 bao ggo M6i bao c3n n{ng lAn lugt ld 15k9, 19k9,20k9, 21kg,27kg vA.28kg Crla hdng da bet5 bao gqo cho 2 nguoi kh6ch S6 ki-lO-gam gqo ngudith0 nhet mua g5p d6i sO ki-16-gam nguoithrt hai H6i:
a) Cdn lai bao g4o ndo chua b6n?
b) Ngudi khAch thu nh6t mua nh0ng bao gqo niro? Nguoi kh6ch thu hai mua nh0ng bao gao ndo? Htdng gidi:
Nli gqi s6 ki-16-gam crla-ngudi thf hai mua lA phAn thi s6 ki-16-gam c&a ngudi tho nnAt ta phAn,
ci hai ngirOi la phAln V?y t6ng s6 ki-t6-gam gAo da Uan
ta m0t s6 chia tr5t cno 3 di6m
Trons diy s6 tS, t 9, 20, 21, 27, 28,cdc s6 chia het cho ld: 15, 21, 27 Vay 6 cAn tim s6 cdn l4i, t6ng s6 ndo chia hdt cho D6 la 19, 20'
Vay Oao chua b6n i-a Oao ning'28kg, di6m
56 fito-gam gqo cl5 b6n: 15 + 19 + 20+ 21 + 27 = 02 (kg)
S6 fitO-iam iqo nguoi thu hai mua: 102 : = 34 (kg) 0,5 diQm
Ngudi th0 nai mua-Z bao gao: 15, 19 0,5 di6-m Sd ti-tO-gam gqo ngudi thi?.nnAt mual 34 x = 68 (kg) 0,5 di6-m
Ngudi thr? ntr-5t mul 3 bao g4o: 20,21 , 27 0,5 (li6m Bi6u di€m:
- Ldi giii hEp ll, tinh dring k6t qui, duEc di6m t6i da crla phAn tt6
- H1i sinh ldm cdch khdc, tinh dhng k6t qud:
(3)BAf 4: G di6m)
Hinh chft nhit ABCD c6 chi6u dai AB ld 4cm, chiAu rQng BC lA 3cm
a) Hiy v6 hai hinh ch0 nh?t c6 cirng diQn tich vdi
hinh ch0 hf'at neCO nhung c6 chi6u ddi, chi6u rQng kh6c vdi hinh chir nhit ABCD
b) TrEn cAnh AB l6y di6m P, tr6n cqnh -CD lAy di6m'Q cho np = iQ Gqi M lA trung di6m c0a BC Tinh di€n tich PBCQ, di€n tich PMQ
Htdng gidi:
i) -oign tich hinh cho nh?t ABCD: 4 x =.12 (cmz)
C6c canh crja hinh chir nh?t c6 di€n tich bing 12cm2 co th6 la: 6cm vd 2cm; 12cm vdr 1cm (VE 2 hinh)
b) Hinh thang PBCQ c6 t6ng hai d6y bing chidu ddi hinh ch0 nhat n6n dign tich bing:
(ax3):2= 6(cm')
Haihinh tam qidc PBM vA MCQ c6:
Chi6u cao c0ig oang % chi\u rQng hinh ch0 nh?t ABCD: 3'.2 = 1,5 (cm)
T6ng 2 Aai pb, CQ bing chiAu dAi hinh ch0 nh?t ABCD ndn t6ng dien tich hai tam gi6c PBM vA MCQ bing:
(1,5x4):2= 3(cm')
V?y dign tich PMQ li: 6 - 3 = (cm')
Bi6:u di€m:
- Ldi giii hgp li, tinh dtng di€n tich hinh ch0nh?t ABCD la 12cm' duqc 0,5 di6m - vc i ninn 6ho nhit voi t<icn thudc ctung, m6i hinh duEc 0,75 di6m
- Ldi giii hgp li, tinn e0ng dien tich hinhlhang PBCQ ld 6cm2 dusc 1,5 di6m' - Ldi diai h;'e li: tinh cldn! diin tich hinh tam giac nMo ld 3cm2 duqc 1,5 di6m
Hgc sinh titm cdch khdc dtng vdn xem x6ttinh di s6 di6m
Di6m trinh biy, chCr viiit: t ei6m
- Trinh bdry 16 rirng, ch0 vi6t dpp, kh6ng boi xod
- Trinh bdy 16 rdng, ch0 vi6t Clgp, c6 boi xo6 - Trinh bdy 16 rdrng, khong b6ixo6
- Trinh bAy 16 rdng
DiAm dwgc tinh ililn 0,25.
: 1,00 di6m.
: 0,50 di6m. : 0,25 di6m.