["TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 a. 3 b. 4 c. 5 Ph\u1ea7n 2: Th\u1ef1c h\u00e0nh 1\/ Vi\u1ebft c\u00e1c s\u1ed1 5 , 9 , 2 , 7 , 4: a. Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn: b\u00e9: ........................................................................................... \u0111\u1ebfn b. Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn ........................................................................................... 3\/ T\u00ednh: 2 + 1 + 1 = .................. 2 + 2 + 1 = .................. 3 + 0 + 2 = .................. 4\/ T\u00ednh: 2 4 32 + + ++ 3 0 12 \u0110\u1ec0 39 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 B\u00e0i 1 : S\u1ed1 ? Th\u1eddi gian : 40 ph\u00fat a\/ 01 3 5 b \/ S\u1ed1 l\u1edbn nh\u1ea5t c\u00f3 m\u1ed9t ch\u1eef s\u1ed1 l\u00e0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026. S\u1ed1 b\u00e9 nh\u1ea5t c\u00f3 m\u1ed9t ch\u1eef s\u1ed1 l\u00e0 \u2026\u2026\u2026\u2026\u2026\u2026........ B\u00e0i 2 : T\u00ednh : 1 + 2 = ............... 1 + 2 + 1 = ............ 0 + 4 = ................ 1 + 4 + 0 = \u2026\u2026\u2026 B\u00e0i 3 : Vi\u1ebft c\u00e1c s\u1ed1 5 , 8 , 2 , 3 : a\/ Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn:............................................................................ b\/Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9: ........................................................................... 50 S\u1ed1 ?","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 B\u00e0i 4 : 2+2 = 5 =4+ + 1= 3 B\u00e0i 5 : \u0110i\u1ec1n d\u1ea5u : > , < , = v\u00e0o ch\u1ed7 ch\u1ea5m . > ? 2 + 0 \u2026\u2026.. 0 5 \u2026\u2026. 4 + 1 < 4 + 1 \u2026. \u2026 5 = 2 + 0 \u2026\u2026.. 2 B\u00e0i 6 : S\u1ed1? H\u00ecnh tam gia\u00f9c Ba\u00f8i 7 : Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: \u0110\u1ec0 40 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat Ph\u1ea7n I : Khoanh tr\u00f2n v\u00e0o ch\u1eef \u0111\u1eb7t tr\u01b0\u1edbc k\u1ebft u\u1ea3 \u0111\u00fang 1. S\u1ed1 b\u00e9 nh\u1ea5t c\u00f3 1 ch\u1eef s\u1ed1 l\u00e0: A.0 B.9 C.1 D.2 2. S\u1ed1 b\u00e9 nh\u1ea5t c\u00f3 2 ch\u1eef s\u1ed1 gi\u1ed1ng nhau l\u00e0: A. 99 B. 22 C.10 D. 11 3.S\u1ed1 li\u1ec1n tr\u01b0\u1edbc c\u1ee7a 90 l\u00e0: A.88 B.89 C.80 D.91 4. C\u00f3 bao nhi\u00eau s\u1ed1 c\u00f3 m\u1ed9t ch\u1eef s\u1ed1? 51","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 A. 8 B. 9 C. 10 D. 11 5. Cho d\u00e3y s\u1ed1 sau: 11; 13; 15; .......; .........; 21. Hai s\u1ed1 c\u00f2n thi\u1ebfu l\u00e0: A. 15; 17 B. 17; 19 C.19; 20 D. 21; 23 6. 10 cm = .........dm. S\u1ed1 th\u00edch h\u1ee3p \u0111i\u1ec1n v\u00e0o ch\u1ed7 ch\u1ea5m l\u00e0: A. 10 B. 12 C. 1 D. 100 7. Trong h\u00ecnh v\u1ebd b\u00ean c\u00f3 ........... h\u00ecnh t\u1ee9 gi\u00e1c. S\u1ed1 th\u00edch h\u1ee3p \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0: A. 4 B. 6 C.9 D. 8 8. N\u0103m nay em 8 tu\u1ed5i, hai n\u0103m n\u1eefa tu\u1ed5i em s\u1ebd l\u00e0: A. 9 tu\u1ed5i B. 10 tu\u1ed5i C. 11 tu\u1ed5i D. 12 tu\u1ed5i Ph\u1ea7n 2: b\/ 5 v\u00e0 21 . B\u00e0i 1 \/ \u0110\u1eb7t t\u00ednh r\u1ed3i t\u00ednh t\u1ed5ng, bi\u1ebft c\u00e1c s\u1ed1 h\u1ea1ng l\u00e0: a\/ 43 v\u00e0 25 B\u00e0i 2 \/ \u0110\u1eb7t t\u00ednh r\u1ed3i t\u00ednh hi\u1ec7u, bi\u1ebft s\u1ed1 b\u1ecb tr\u1eeb v\u00e0 s\u1ed1 tr\u1eeb l\u1ea7n l\u01b0\u1ee3t l\u00e0: a\/ 84 v\u00e0 31 b\/ 59 v\u00e0 9 B\u00e0i 3\/ T\u00ednh 8dm + 7 dm \u2013 5dm = .............. 19cm \u2013 10cm = .............. 16l + 4l \u2013 5 l = .............. 55kg + 4kg = ............... B\u00e0i 4\/ Gi\u1ea3i to\u00e1n : Tu\u1ea5n c\u00e2n n\u1eb7ng 38 kg. Minh c\u00e2n n\u1eb7ng h\u01a1n Tu\u1ea5n 6 kg. H\u1ecfi Minh c\u00e2n n\u1eb7ng bao nhi\u00eau ki-l\u00f4-gam? \u0110\u1ec0 41 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat I\/ Tr\u1eafc nghi\u1ec7m: C\u00e2u 1: S\u1ed1 b\u00e9 nh\u1ea5t c\u00f3 m\u1ed9t ch\u1eef s\u1ed1 l\u00e0: A. 1 B. 2 C. 0 D. 3 C\u00e2u 2: S\u1ed1 l\u1edbn nh\u1ea5t c\u00f3 m\u1ed9t ch\u1eef s\u1ed1: A. 8 B. 7 C. 9 D. 6 C\u00e2u 3: S\u1ed1 li\u1ec1n tr\u01b0\u1edbc s\u1ed1 8 l\u00e0: A. 7 B. 9 C. 6 D. 5 52","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 C\u00e2u 4: S\u1ed1 li\u1ec1n sau s\u1ed1 7 l\u00e0: A. 8 B. 10 C. 9 D. 6 C\u00e2u 5: S\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 9< A. 8 B. 10 C. 7 D. 6 C\u00e2u 6: D\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0:7 9 A. < B. > C. = C\u00e2u 7: K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh l\u00e0: 2+1+2 =\u2026\u2026 A. 5 B. 6 C. 4 D. 3 C\u00e2u 8: D\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 1+3 3+1 A. < B. > C. = C\u00e2u 9: K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh l\u00e0: 2+3=.... A. 3 B. 4 C. 5 D.6 C\u00e2u 10: H\u00ecnh v\u1ebd b n c\u00f3 bao nhi u h\u00ecnh tam gi\u00e1c? A. 2 B. 3 C. 4 D. 1 Th\u1ef1c h\u00e0nh: C\u00e2u 1: Vi\u1ebft s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng: 1 5 03 7 10 C\u00e2u 2: T\u00ednh: 2 + 1 + 2 =\u2026\u2026 2 4 2 + 2 =\u2026\u2026 + + 3 + 2 =\u2026\u2026 1 1 \u2026\u2026. \u2026. C\u00e2u 3: \u0110i\u1ec1n d\u1ea5u <, > , = th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng: 65; 9 10 ; 4 4 C\u00e2u 4: \u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng: 5+ =5 ; 4+1= 1+ \u0110\u1ec0 42 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat B\u00e0i 1: S\u1ed1? \u25b2\u25b2\u25b2 \u266a\u266a\u266a\u266a\u266a \u2642\u2642\u2642\u2642 \u2663\u2663\u2663\u2663 \u263c\u263c\u263c \u25b2 \u2642\u2642\u2642 \u263c\u263c\u263c 53","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 \u266a \u2663\u2663\u2663\u2663 \u263c\u263c\u263c aB\u00e0i 2: Vi\u1ebft s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng: 14 54 B\u00e0i 3 :T\u00ednh : 0 3 3 2 + + + + 3 2 0 3 B\u00e0i 4 : T\u00ednh : 2+1+2= 3 +1 + 1 = 2 +1 + 1 = 5 + 0 ...... 2 + 3 B\u00e0i 5 : > < = ? 2 + 3 ...... 5 2 + 2 ...... 2 + 1 B\u00e0i 6 : Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p : v\u00e0 B\u00e0i 7 : H\u00ecnh b\u00ean : a) C\u00f3 m\u1ea5y h\u00ecnh tam gi\u00e1c ? \u0110\u1ec0 43 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I 54","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat 1. S\u1ed1 ? \uf0a1 \uf0bf\uf0bf\uf0bf \uf0b6\uf0b6\uf0b6 \uf0b6\uf0b6 \uf0a8\uf0a8 \uf07b\uf07b \uf0a8\uf0a8 2. Vi\u1ebft s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng: 4 0 51 3. ? 2;2 4;3 2;4 5 > 22 4 <2 +1 +3 +1 = 4.T\u00ednh: 1 +3 ... ... ... ... 5. Vi\u1ebft s\u1ed1 v\u00e0 ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: \uf06c \uf06c\uf06c \uf06c\uf06c \uf06c\uf06c \uf06c 6. T\u00ednh: 2 + 2 = ............ 1 + 4 = ............ 55","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 3 + 2 = ............ 5 + 0 = ............ \u0110\u1ec0 44 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat 1. Vi\u1ebft: Vi\u1ebft c\u00e1c s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 10 : ....................................................................................... 2. T\u00ednh : a) 3 221 1124 ........ ....... ....... ...... b) 2+0+1 = ..........; 1+3+1 =..............; 5+0 =...............; 2+3= .............. 3. Vi\u1ebft c\u00e1c s\u1ed1 : 5; 6; 2; 3; 7 a) Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn:..................................................................................... b) Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9: ................................................................................... 4. Khoanh v\u00e0o ch\u1eef \u0111\u1eb7t tr\u01b0\u1edbc k\u1ebft u\u1ea3 \u0111\u00fang: H\u00ecnh d\u01b0\u1edbi \u0111\u00e2y c\u00f3 m\u1ea5y h\u00ecnh tam gi\u00e1c? A. 1 h\u00ecnh C. 3 h\u00ecnh B. 2 h\u00ecnh D. 4 h\u00ecnh 5. S\u1ed1 ? ......+ 3 = 3; 3+.......= 5 ......+ 1= 2 6. 3+2......1+2 2+1......1+2 > ? 2+3.......5 < 2+2.......5 = 7. Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: ** * ** 56 ** **","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 \u0110\u1ec0 45 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat 1. N\u1ed1i theo m\u1eabu: \uf0cb\uf0cb\uf0cb\uf0cb\uf0cb XXXX \uf0cb\uf0cb\uf0cb\uf0cb\uf0cb XXXX 10 \uf025\uf025\uf025\uf025 78 \uf025\uf025\uf025\uf025 \uf022\uf022\uf022\uf022 \uf022\uf022\uf022 2. Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: b) a) \uf028 \uf028\uf028\uf028\uf028 \uf026 \uf026\uf026 3. So\u00e1 ? 1 + ... = 1 2 + ... = 4 +1 1 +4 2 1 + 2 + 1 = ... 4. T\u00ednh: 2 + 1 + 2 = ... 10 \u2026 0 5. > 5\u2026 8 7\u20265 <? 9\u2026 2 4 + 0 ... 2 + 3 = 5 ... 2 + 1 57","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 6.H\u00ecnh b\u00ean c\u00f3: \u2026h\u00ecnh vu\u00f4ng. \u0110\u1ec0 46 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat C\u00e2u 1: T\u00ednh nh\u1ea9m: 5+8=\u2026 7+0 =\u2026 3+9 =\u2026 9+6=\u2026 8+6=\u2026 18 + 5 = \u2026 6+7 =\u2026 7+7=\u2026 6+5=\u2026 8+7 =\u2026 9+5 =\u2026 9+8=\u2026 C\u00e2u 2: \u0110\u00e1nh d\u1ea5u X v\u00e0o \u00f4 \u0111\u00fang: b) 19 \u2013 12 - 4 = 3 a) 10 + 9 \u2013 5 = 13 19 \u2013 12 - 4 = 4 10 + 9 \u2013 5 = 14 19 \u2013 12 - 4 = 5 10 + 9 \u2013 5 = 15 c) 10cm = 1dm d) 6dm = 6cm 10cm = 100dm 6dm = 60 cm C\u00e2u 3: Trong h\u00ecnh b\u00ean: a\/ C\u00f3\u2026.. h\u00ecnh tam gi\u00e1c b\/ C\u00f3\u2026.. H\u00ecnh t\u1ee9 gi\u00e1c C\u00e2u 4: \u0110\u1eb7t t\u00ednh r\u1ed3i t\u00ednh : 38 + 56 ; 69 + 17 ; 45 + 39 ; 9 + 64 \u2026\u2026 \u2026\u2026 \u2026\u2026 \u2026\u2026 \u2026\u2026 \u2026\u2026 \u2026\u2026 \u2026\u2026 \u2026\u2026 \u2026\u2026 \u2026\u2026 \u2026 C\u00e2u 5: 58","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 M\u1eb9 h\u00e1i \u0111\u01b0\u1ee3c 35 qu\u1ea3 b\u01b0\u1edfi, ch\u1ecb h\u00e1i h\u01a1n m\u1eb9 18 qu\u1ea3 b\u01b0\u1edfi . H\u1ecfi ch\u1ecb h\u00e1i \u0111\u01b0\u1ee3c bao nhi u qu\u1ea3 b\u01b0\u1edfi ? C\u00e2u 6: D\u00f9ng th\u01b0\u1edbc v\u00e0 b\u00fat n\u1ed1i c\u00e1c \u0111i\u1ec3m \u0111\u1ec3 c\u00f3 1h\u00ecnh ch\u1eef nh\u1eadt: AB \u2219\u2219 \u2219E \u2219\u2219 C D \u0110\u1ec0 47 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I C\u00e2u 1: S\u1ed1? M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat \uf04a\uf04a \uf059\uf059 \uf052 \uf052\uf052 \uf04a \uf059 \uf052 \uf04a\uf04a \uf04a \uf059\uf059 \uf052\uf052 \uf059\uf059 \uf052 \uf052\uf052 \uf04a\uf04a \uf052 C\u00e2u 2: Vi\u1ebft s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng: 1 57 6 9 47 555 5 10 8 C\u00e2u 3: Vi\u1ebft c\u00e1c s\u1ed1 3, 6, 10 ,7, 9 . a) Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn b) Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9. C\u00e2u 4 :< ? 3 + 2 5 ; 2+2 5+0 > 59 =","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 4+1 3+1; 5+0 0+4 C\u00e2u 5: T\u00ednh 5 3 24 0 2 11 \u2026.. \u2026.. \u2026\u2026. \u2026... C\u00e2u 6 : H\u00ecnh b n c\u00f3: \u2026. h\u00ecnh tam gi\u00e1c C\u00e2u 7: Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: \u0110\u1ec0 48 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I B\u00e0i 1: S\u1ed1? M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat 1 47 5 B\u00e0i 2: T\u00ednh 3 + 0 + 1 = .......... 2 + 1 + 1 = .......... 1 + 4 = .......... 4 + 1 + 0 = .......... 2 + 2 = .......... 4 + 0 = .......... 4 4+0 3+1 3+0 B\u00e0i 3: > <? 2 2 + 3 = 5 2+2 3 1+1 2+3 4+0 B\u00e0i 4: Vi\u1ebft c\u00e1c s\u1ed1 0, 1, 7, 3, 4: a. Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn:.............................................................................................. b. Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9:.............................................................................................. 60","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 B\u00e0i 5: Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: B\u00e0i 6: H\u00ecnh b\u00ean c\u00f3: h\u00ecnh vu\u00f4ng. \u0110\u1ec0 49 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat 1\/ S\u00e8 ? 3 5 8 10 2 Vi\u00d5t c\u00b8c s\u00e8 : 2 , 10 , 6 , 9 *Theo th\u00f8 t\u00f9 t\u00f5 b\u00d0 \u00ae\u00d5n l\u00edn : *Theo th\u00f8 t\u00f9 t\u00f5 l\u00edn \u00ae\u00d5n b\u00d0 : 2\/ T\u00ddnh : 1 \u2026\u2026 5 \u2026\u2026. 2 4 2 2 +1 +1 + +3 +0 \u2026\u2026 \u2026\u2026 +2 0 61 \u2026.\u2026 \u2026\u2026","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 3\/ T\u00ddnh : 3+2 = \u2026..\u2026 1 + 0 + 4 = \u2026\u2026 2 + 1 + 1 = \u2026\u2026. 2 + 1 + 2 = \u2026..\u2026 4\/ \u00a7i\u00d2n d\u00cau : > , < , = 2+3 1+4 3+0 4 57 2+2 2+1 0+2 2+3 75 5\/ S\u00e8 : 6\/ Vi\u00d5t ph\u00d0p t\u00ddnh th\u00ddch h\u00eep v\u00edi h\u00d7nh v\u00cf b\u00aan \u0110\u1ec0 50 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 B\u00e0i 1: \u0110i\u1ec1n s\u1ed1 0 Th\u1eddi gian : 40 ph\u00fat 2 45 98 32 62","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 97 4 78 B\u00e0i 2: Vi\u1ebft c\u00e1c s\u1ed1 9 , 7 , 1 , 3 , 5 , 6: \uf0b7 Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn: .......................................................... \uf0b7 Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9: .......................................................... B\u00e0i 3: \u25a1>? 0 1 \u25a13 9 \u25a18 5 \u25a17 7 \u25a14 8 < \u25a1= 10 6 B\u00e0i 4 : T\u00ednh 3 + 2 = ............. + + 4 + 0 = ............. .......... ........... 2 + 3 = ............. 0 + 3 = ............. B\u00e0i 5: Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p \uf093\uf093 \uf093\uf093\uf093 B\u00e0i 6 : S\u1ed1? H\u00ecnh d\u01b0\u1edbi \u0111\u00e2y c\u00f3 : 63","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 C\u00f3 ................. h\u00ecnh tam gi\u00e1c C\u00f3 ................. h\u00ecnh vu\u00f4ng \u0110\u1ec0 51 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I 64","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat 1.S\u1ed1? 35 8 74 2.Vi\u1ebft c\u00b8c s\u00e8 sau 9 , 5 , 7 , 2 . a. Theo th\u00f8 t\u00f9 t\u00f5 l\u00edn \u00ae\u00d5n b\u00d0: .......................................................... .......... b. Theo th\u00f8 t\u00f9 t\u00f5 b\u00d0 \u00ae\u00d5n l\u00edn: .......................................................... .......... 3.T\u00ddnh : 22 4 +3 +2 +1 .......................................................................................................................................... 4.. >79 2+2 5 <? =09 1+3 4 5.Hinh d\u01b0\u1edbi c\u00f3: \u2026 h\u00d7nh tam gi\u00b8c . 6.S\u1ed1? \u206d+4=4 2 + \u206d =2 3 + \u206d=5 \u206d +4 = 5 7.Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: 65","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 \u0110\u1ec0 52 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat 1) > < = ? 3 ....... 2 8 ........ 6 9 ....... 9 6 ....... 5 + 0 7 ....... 7 5 ....... 4 + 1 4 ....... 5 8 ........ 9 2) Vi\u1ebft s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng: 14 8 10 1 5 + 0 = ....... 3) T\u00ednh: 0 + 5 = ....... 1 + 4 = ....... 2+1+2 2+0+2= =....... 0 4 + 1 = ....... +5 ............ ...... 4) T\u00ednh: 2 +4 1 3 1 +3 0 +2 +2 +4 ............ ............ .......... ........... ............. 5) Khoanh v\u00e0o s\u1ed1 l\u1edbn nh\u1ea5t a) 4 , 3 , 5 b) 9 , 10, 8 6) Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: 66","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 \u0110\u1ec0 53 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat I. Tr\u1eafc nghi\u1ec7m : Khoanh tr\u00f2n v\u00e0o ch\u1eef c\u00e1i \u0111\u1eb7t tr\u01b0\u1edbc c\u00e2u tr\u1ea3 l\u1eddi \u0111\u00fang. 1. Trong c\u00e1c s\u1ed1 2,8,5,9,1. S\u1ed1 b\u00e9 nh\u1ea5t l\u00e0 : a. 1 b. 2 c. 9 2. Trong c\u00e1c s\u1ed1 10, 4, 7, 6, 9. S\u1ed1 l\u1edbn nh\u1ea5t l\u00e0 : a. 9 b. 10 c. 4 3. 2 + 3 = \u2026. S\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 ch\u1ea5m l\u00e0 : a. 4 b. 5 c.3 4. 8\u2026..5. D\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 ch\u1ea5m l\u00e0 : a. < b. > c. = 5. 1 + 4 S\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 : a. 5 b. 4 c.3 II. Th\u1ef1c h\u00e0nh C\u00e2u 1 : Vi\u1ebft s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng \/ 14 8 C\u00e2u 2 : T\u00ednh : 2 + 2 = \u2026\u2026\u2026\u2026.. 24 1 + 2 + 2 = \u2026\u2026\u2026\u2026 ++ 31 C\u00e2u 3 : > 1 + 2 \u2026\u2026. 3 2 \u2026\u20262 + 0 4 + 1 \u2026\u2026.4 <? = \u01a0 67 = =","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 C\u00e2u 4:S\u1ed1? + 2=3 4+ =5 C\u00e2u 5: Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: C\u00e2u 6: H\u00ecnh b\u00ean c\u00f3 : \u2026\u2026 h\u00ecnh tam gi\u00e1c. \u0110\u1ec0 54 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat C\u00e2u 1: > ? a\/ 10.\u2026..7 4 ...... 5 < b\/ 8\u2026.4 + 4 1 + 3 .\u2026. 1 + 2 = C\u00e2u 2: Vi\u1ebft s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng: a\/ 1 46 b\/ 64 2 C\u00e2u 3: T\u00ednh : a\/ 2 + 1 + 2 = .......... 1 + 1 + 1 = ........... b\/ 1 + 2 + 1 = ......... 2 + 0 + 1 = ........... C\u00e2u 4: T\u00ednh: - 2 + 2 + 3 - 3 1 1 2 1 ........ ........ ........ ....... C\u00e2u 5: S\u1ed1 ? 68","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 a\/ 2 + \u2026.. = 4 3 + \u2026.. = 5 5 = \u2026.. + 4 b\/ 4 = 1 + \u2026\u2026 C\u00e2u 6: Vi\u1ebft c\u00e1c s\u1ed1 5, 2, 4, 10, 7 : a\/ Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn: \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. b\/ Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9: \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. C\u00e2u 7: Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: \uf0e4\uf0e4 \uf0e4 \uf0e4\uf0e4 \u0110\u1ec0 55 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat 1. 2 3 35 54 7 7 > <? 2. S\u1ed1=? < 2 < 3 > 5 6= 3. T\u00ednh: 1+3 = 3+ 0= 4+1= 0+4= 2+2 = 2+ 1= 5+0= 3+2= 4. Vi\u1ebft s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng: 1+ .= 4 5+ .= 5 + 3 =5 +2 = 4 5. Trong c\u00e1c s\u1ed1 t\u1eeb 0 \u0111\u1ebfn 10: 69","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 a. S\u1ed1 b\u00e9 nh\u1ea5t l\u00e0:\u2026.. b. S\u1ed1 l\u1edbn nh\u1ea5t l\u00e0:\u2026. 6. H\u00ecnh d\u01b0\u1edbi : C\u00f3 \u2026 h\u00ecnh tam gi\u00e1c 7.Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: \u0110\u1ec0 56 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat 1. Vi\u1ebft: Vi\u1ebft c\u00e1c s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 10 : ....................................................................................... 2. T\u00ednh : a) 1 211 3224 ........ ....... ....... ...... b) 2+1+1 = ..........; 2+2+1 =..............; 5+0 =...............; 3+3= .............. 3. Vi\u1ebft c\u00e1c s\u1ed1 : 0; 3; 5; 1; 6 a) Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn:..................................................................................... b) Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9: ................................................................................... 4.H\u00ecnh d\u01b0\u1edbi \u0111\u00e2y c\u00f3 : ...h\u00ecnh tam gi\u00e1c 70","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 5.\u0110i\u1ec1n S\u1ed1 ? . ......+ 3 = 4; 3+.......= 3 ......+ 1= 2 6. Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: ** * ** ** ** \uf0b7 \u0110\u1ec0 57 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat I. PH\u1ea6N TR\u1eaeC NGHI\u1ec6M B\u00e0i 1:S\u1ed1? C\u00e2u a: 1 35 Trong c\u00e1c s\u1ed1 tr n: S\u1ed1 l\u1edbn nh\u1ea5t l\u00e0: C\u00e2u b: S\u1ed1 b\u00e9 nh\u1ea5t l\u00e0: C\u00e2u c 3+2= 3+1= C\u00e2u d: 2+3 = 0+5= B\u00e0i 2: Khoanh tr\u00f2n v\u00e0o ch\u1eef c\u00e1i tr\u01b0\u1edbc \u00fd tr\u1ea3 l\u1eddi \u0111\u00fang: C\u00e2u a: 5 + = 5 A. 1 B. 0 C. 5 D. 2 71","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 C\u00e2u b: S\u1ed1 ng\u00f4i sao c\u00f3 trong h\u00ecnh v\u1ebd b n l\u00e0: A. 5 B. 7 C. 8 D. 9 B\u00e0i 3: > 2 + 2 ...... 2 + 1 5 + 0 ...... 2 + 3 <? = II. PH\u1ea6N T\u1ef0 LU\u1eacN B\u00e0i 1: T\u00ednh 33 0 1 ++ + + .1 2 4 2 \u2026\u2026. \u2026\u2026.. \u2026\u2026\u2026 \u2026\u2026\u2026 B\u00e0i 2: Vi\u1ebft c\u00e1c s\u1ed1 2; 6; 4; 0; 8. a. Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9: \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. b. Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn: \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. B\u00e0i 3: Vi\u1ebft ph\u00e9p t\u00ednh c\u1ed9ng th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng. \u0110\u1ec0 58 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat C\u00e2u 1: Vi\u1ebft c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng: 03 8 10 7 4 0 C\u00e2u2: 72","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 > 2\uf0a8 5 1+1 \uf0a8 2 <? 9\uf0a8 6 5+0 \uf0a8 8 = 3 + 2 =............. C\u00e2u 3: T\u00ednh: 1 + 1 + 2 =............. 2 + 1 =.............. 4 + 0 =............. C\u00e2u 4: Vi\u1ebft c\u00e1c s\u1ed1 10, 7, 1, 3, 5: a. Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn: .............................................................................................................................. a. Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9: .............................................................................................................................. C\u00e2u 5: H\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y : C\u00f3....... h\u00ecnh tam gi\u00e1c C\u00e2u 6:Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: \u0110\u1ec0 59 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 B\u00e0i 1: S\u1ed1? Th\u1eddi gian : 40 ph\u00fat 35 B\u00e0i 2: Vi\u1ebft c\u00e1c s\u1ed1 8 , 3, 5, 6, 10 theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9: 73","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 B\u00e0i 3: Khoanh v\u00e0o s\u1ed1 b\u00e9 nh\u1ea5t: 5; 4; 7; 2; 9 B\u00e0i 4: C\u00e1c s\u1ed1 b\u00e9 h\u01a1n 7 l\u00e0: \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. B\u00e0i 5: > 4+1 1+4 5+ 0 1 + 3 <? 2+2 5+0 2 +3 5 = B\u00e0i 6: T\u00ednh: 2 + 0 + 2 = \u2026\u2026\u2026\u2026\u2026.. 2 + 1 + 2 = \u2026\u2026\u2026\u2026\u2026\u2026\u2026.. B\u00e0i 7: H\u00ecnh v\u1ebd b n c\u00f3: \u2026\u2026. h\u00ecnh tam gi\u00e1c B\u00e0i 8: Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: \u0110\u1ec0 60 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 1\/ T\u00ednh: Th\u1eddi gian : 40 ph\u00fat 74","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 2 + 3 =\u2026\u2026\u2026\u2026.. 2 + 0 + 1 = \u2026\u2026\u2026\u2026.. 1+ 3 = \u2026\u2026\u2026\u2026. 1 + 3 + 1 = \u2026\u2026\u2026\u2026.. 2\/ T\u00ednh: 4 12 2 +0 +3 + 3 +2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. 3\/ S\u1ed1? 14 10 7 3 4\/ S\u1ed1? : 2+ = 4 + 4+ =4 4 +3 =5 1 =4 5\/ 3 + 2 \u2026\u2026 > 1 \u2026.. 5 4 \u2026\u2026 4 + 1 <? 2 = \u2026\u2026 2 + 0 3+ 6\/ Vi\u1ebft c\u00e1c s\u1ed1 3 , 7 , 8 , 4 , 6: a\/ Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn: \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 b\/ Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9: \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 7\/ Vi\u00e3\u00fat phe\u00efp t\u00eanh th\u00each h\u00e5\u00fcp: v\u00e0 75","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 \u0110\u1ec0 61 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat Ba\u00eci 1: >0 1 10 5 4 < 2+2 2 +1 2 86 9 1+ 2 =? 7 Ba\u00eci 2: Khoanh v\u00e0o s\u1ed1 l\u1edbn nh\u1ea5t: 6;3;5;9 Ba\u00eci 3: S\u1ed1 ? 1 3 5 7 + Ba\u00eci 4: T\u00eanh: 5 + a. 2 4 0 3 + + 2 1 b. 1 + 2 + 1 = ............................... ; 3 + 2 + 0 = ......................... Ba\u00eci 5:Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p 76","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 \u0110\u1ec0 62 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat * B\u00e0i 1: S\u1ed1 ? 0 ...... ...... 3 ...... ...... ...... 7 ...... ...... 10 * B\u00e0i 2: > ; < ; = ? > 0 ...... 3 6 ...... 5 4 ...... 1 + 3 < 8 ...... 7 2 ...... 2 2 + 3 ...... 3 = * B\u00e0i 3: T\u00ednh ? 1 + 4 = ...... 2 2 + 1 = ...... 3 + 2 = ...... +3 01 ......... ++ 43 ......... ......... * B\u00e0i 4: Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p \uf07b\uf07b \uf07b\uf07b * B\u00e0i 5: H\u00ecnh v\u1ebd b n c\u00f3 ......... h\u00ecnh vu\u00f4ng 77","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 *B\u00e0i 6:S\u1ed1? 3+ =3 +1=2 4 + =5 +5 = 5 \u0110\u1ec0 63 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat B\u00e0i 1: S\u1ed1 \uf07b\uf07b\uf07b \uf07b a. \uf07b\uf07b\uf07b \uf04a\uf04a\uf04a\uf04a \uf04a\uf04a\uf04a b. 13 975 B\u00e0i 2:T\u00ednh 2 4 3 +2 +0 +2 a. 1 +3 ; 2 + 1 = \u2026\u2026\u2026\u2026 ; 1 +1 + 2 =\u2026\u2026\u2026. b. 1 + 2 = \u2026\u2026\u2026\u2026\u2026. 3 + 0 + 2 = \u2026\u2026\u2026\u2026.. 78 B\u00e0i 3.: a. (1\u0111i\u1ec3m) > < =","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 ? 2 + 1 \u2026\u2026.. 3 ; 4 + 0\u2026\u2026.. 5 b. S\u1ed1? \uf0a8 >8 \uf0a8 <3 B\u00e0i 4: a. Vi\u1ebft c\u00e1c s\u1ed1 0,5,2,6,10 theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn: b. Vi\u1ebft c\u00e1c s\u1ed1 8,4,2,6,10 theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9 8; 4; 2; 6; 10 B\u00e0i 5: Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p \uf0ff\uf0ff \uf0ff\uf0ff\uf0ff B\u00e0i 6: H\u00ecnh d\u01b0\u1edbi c\u00f3 m\u1ea5y h\u00ecnh tam gi\u00e1c C\u00f3\u2026\u2026\u2026\u2026.h\u00ecnh tam gi\u00e1c 79","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 \u0110\u1ec0 64 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 Th\u1eddi gian : 40 ph\u00fat B\u00e0i 1:N\u1ed1i theo m\u1eabu: 34 5 67 8 B\u00e0i 2: X\u1ebfp c\u00e1c s\u1ed11 , 5 , 9 , 3 . 7 theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn : 80","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 ............................................................................................................................................ B\u00e0i 3. T\u00ednh : a) 2 + 2 = .......... 3 + 2 = ........... 3 + 0 = ......... 3 + 1 = ......... b) + 21 1 5 + + + 34 2 0 ......... ......... ......... ........ c) 2 + 1 +1 = .......... 2 + 3 + 0 = .......... B\u00e0i 4: 0 + 3 \u2026\u2026 2 + 1 1 + 2 \u2026\u2026 3 + 1 > ? 3 + 2 \u2026\u2026 5 < = 1 + 3 \u2026\u2026 5 B\u00e0i 5: Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p : B\u00e0i 6: S\u1ed1? 3+2= + \u0110\u1ec0 65 KI\u1ec2M TRA \u0110\u1ecaNH K\u00cc GI\u1eeeA H\u1eccC K\u00cc I M\u00f4n: To\u00e1n \u2013 L\u1edbp 1 B\u00e0i 1: a. S\u1ed1 ? Th\u1eddi gian : 40 ph\u00fat 81","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 b. 3 8 8 7 6+0 0 +5 >4 <? =5 B\u00e0i 2: Vi\u1ebft c\u00e1c s\u1ed1 8, 5, 2,7, 10: a. Theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn: b. Theo th\u1ee9 t\u1ef1 t\u1eeb l\u1edbn \u0111\u1ebfn b\u00e9: B\u00e0i 3: T\u00ednh : 5 2 2 a\/ 3 + + + + 2 0 1 2 \u2026\u2026\u2026.. \u2026\u2026\u2026.. \u2026\u2026\u2026.. \u2026\u2026\u2026.. 2+0+1= b\/ 1 + 3 + 1 = .......... ; 2 + 1 + 0 = ..........; 2 + 2 + 1= .........; ........ B\u00e0i 4: Vi\u1ebft ph\u00e9p t\u00ednh th\u00edch h\u1ee3p: a. B\u00e0i 5: H\u00ecnh? b. C\u00f3 ........... h\u00ecnh tam a.C\u00f3 .......... h\u00ecnh tr\u00f2n gi\u00e1c 82","TUY\u1ec2N T\u1eacP 65 \u0110\u1ec0 KI\u1ec2M TRA GI\u1eeeA H\u1eccC K\u00cc I \u2013 M\u00f4n To\u00e1n l\u1edbp 1 83"]
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