ﻤﺤﺘﻭﻴﺎﺕ ﺍﻹﺭﺴﺎل ﺍﻷﻭل ﻴﺘﻀﻤﻥ ﻫﺫﺍ ﺍﻹﺭﺴﺎل ﺍﻟﻤﻭﺍﻀﻴﻊ ﺍﻟﺘﺎﻟﻴﺔ: • ﻤﺠﻤﻭﻋﺎﺕ ﺍﻷﻋﺩﺍﺩ• ﺍﻷﻋﺩﺍﺩ ﺍﻷﻭﻟﻴﺔ ﻭ ﺍﻟﻘﻭﺍﺴﻡ ﻭ ﺍﻟﻤﻀﺎﻋﻔﺎﺕ )ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ (N • ﺍﻟﻘﻭﻯ • ﺍﻟﺠﺫﻭﺭ ﺍﻟﺘﺭﺒﻴﻌﻴﺔ • ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ • ﺍﻟﻘﻴﻡ ﺍﻟﻤﻘﺭﺒﺔ ﻭ ﺍﻟﺘﻘﺩﻴﺭﺍﺕ • ﺃﻨﺸﻁﺔ ﺒﺎﻟﺤﺎﺴﺒﺔ ﺍﻟﻌﻠﻤﻴﺔ • ﺃﻨﺸﻁﺔ ﺤﻭل ﺍﻟﺠﻤل ﻓﻲ ﺍﻟﺭﻴﺎﻀﻴﺎﺕ ﻭ ﺍﻟﻤﻨﻁﻕ )(1 • ﺤل ﺍﻟﻤﻌﺎﺩﻻﺕ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﺫﺍﺕ ﺍﻟﻤﺠﻬﻭل ﺍﻟﻭﺍﺤﺩ • ﺍﻟﺘﺭﺘﻴﺏ ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ (1) R • ﺍﻟﺘﺭﺘﻴﺏ ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ (2) R • ﻤﺘﺭﺍﺠﺤﺎﺕ ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ R • ﺃﺸﻌﺔ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻌﺎﻟﻡ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ • ﺃﻨﺸﻁﺔ ﺤﻭل ﺍﻟﺠﻤل ﻓﻲ ﺍﻟﺭﻴﺎﻀﻴﺎﺕ ﻭﺍﻟﻤﻨﻁﻕ )(2
ﻤﺠﻤﻭﻋـﺎﺕ ﺍﻷﻋـﺩﺍﺩ ﺍﻟـﻤﺨـﺘﻠﻔـﺔ ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ: -ﺍﻟﺘﻤﻜﻥ ﻤﻥ ﺍﻟﺤﻜﻡ ﻋﻥ ﻋﺩﺩ ﺇﻥ ﻜﺎﻥ ﻴﻨﺘﻤﻲ ﺇﻟﻰ Nﺃﻭ ﺇﻟﻰ Zﺇﻭ ﺇﻟﻰ Dﺃﻭ ﺇﻟﻰ Qﺃﻭ ﺇﻟﻰ.R -ﺘﻨﻅﻴﻡ ﺤﺴﺎﺒﺎﺕ ﻓﻲ Dﻭﻓﻲ Q ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ • ﻤﺠﻤﻭﻋﺎﺕ ﺍﻷﻋﺩﺍﺩ • ﻤﻼﺤﻅﺎﺕ ﻭﺃﻤﺜﻠﺔ • .ﺃﻤﺜﻠﺔ ﻟﺘﻨﻅﻴﻡ ﺤﺴﺎﺒﺎﺕ ﻓﻲ Dﺃﻭ ﻓﻲ Q • ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ • ﺍﻟﺤﻠﻭل
• ﻤﺠﻤﻭﻋﺎﺕ ﺍﻷﻋﺩﺍﺩ : • Nﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻁﺒﻴﻌﻴﺔ ﺃﻱ: }N = {0, 1,2,…. • Zﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ ﺍﻟﻨﺴﺒﻴﺔ ﺃﻱ: }Z = {…,-2,-1,0,1,2,... Dﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻌﺸﺭﻴﺔ ﺃﻱ Dﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻬﺎ ﻋﻠﻰ ﺍﻟﺸﻜل • • n∈N a∈Zﻭ ﺤﻴﺙ q • 10 nﺤﻴﺙ q Qﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻨﺎﻁﻘﺔ ﺃﻱ Qﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﻋﻠﻰ ﺍﻟﺸﻜل b a∈Zﻭb∈N Rﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺤﻘﻴﻘﻴﺔ.
• ﻤﻼﺤﻅﺎﺕ ﻭﺃﻤﺜﻠﺔ : .1ﻟﺩﻴﻨﺎ ﺍﻹﺤﺘﻭﺍﻴﺎﺕN ⊂ Z ⊂ D⊂Q⊂ R : .2ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ﻜل ﻋﺩﺩ ﻋﺸﺭﻱ ﻋﻠﻰ \"ﺸﻜل ﻜﺴﺭﻱ\" ﺃﻭ ﻋﻠﻰ \"ﺍﻟﺸﻜل ﺍﻟﻌﺸﺭﻱ\" ﺒﺎﺴﺘﻌﻤﺎل ﻋﺩﺩ ﻤﻨﺘﻪ ﻤﻥ ﺍﻷﺭﻗﺎﻡ ﺒﻌﺩ ﺍﻟﻔﺎﺼﻠﺔ.73 = 146 ﻤﻨﻪ 73 = 2 × 73 ﻋﺩﺩ ﻋﺸﺭﻱ ﻷﻥ 73 ﻤﺜﻼ :5 101 5 2×5 5 ﻫﻭ 14.6 73 ﺍﻟﻌﺸﺭﻱ ﻟﻠﻌﺩﺩ ﻭﺍﻟﺸﻜل 5 44 = 4.666 ﻟﻴﺱ ﻋﺩﺩﺍ ﻋﺸﺭﻴﺎ ﻷﻥ..... 44 ﺒﻴﻨﻤﺎ 3 3 .3ﺍﻷﻋﺩﺍﺩ ﺍﻟﺤﻘﻴﻘﻴﺔ ﺍﻟﺘﻲ ﻻ ﺘﻨﺘﻤﻲ ﺇﻟﻰ Qﺘﺴﻤﻰ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺼﻤﺎﺀ
• ﺃﻤﺜﻠﺔ ﻟﺘﻨﻅﻴﻡ ﺤﺴﺎﺒﺎﺕ ﻓﻲ Dﺃﻭ ﻓﻲ : Q ﺇﻨﺘﺒﻪ :ﺍﻟﻜﺘﺎﺒﺔ ) (x – yﺘﻌﻨﻲ ) xﻨﺎﻗﺹ (yﻭﺍﻟﻜﺘﺎﺒﺔ ) (x : yﺘﻌﻨﻲ ) xﺘﻘﺴﻴﻡ(y ﻟﻨﺤﺴﺏ ﺍﻷﻋﺩﺍﺩ D, C, B, A : , A = 2 − 3 + 7 − 1 5 4 2 C = [(0.7 )+1 )−(2−3.8)]×(0.5−3 , B = 1 − 3 − 2 +1 3 7 5 7 100 1 0.3 + − 0.03 1 − 1 1 1+D = 3 1 ﻭ C = 1 ÷ + 1 100 4 3 + 0.05 − 1 − 1 3 1− • ﻟﺤﺴﺎﺏ : Aﻴﻤﻜﻥ ﺃﻥ ﻨﻘﻭﻡ ﺒﺘﻭﺤﻴﺩ ﺍﻟﻤﻘﺎﻤﺎﺕ ﺜﻡ ﻨﺠﻤﻊ ﻭﻟﻜﻥ ﻫﻨﺎ ﻤﻥ ﺍﻷﻓﻀل ﺃﻥ ﻨﻼﺤﻅ ﺃﻥ ﺠﻤﻴﻊ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺴﺘﺨﺩﻤﺔ ﻫﻲ ﺃﻋﺩﺍﺩ ﻋﺸﺭﻴﺔ ﻭﻴﻤﻜﻥ ﺍﺴﺘﻌﻤﺎل ﺍﻟﺤﺎﺴﺒﺔ A = 0.4 – 0.75 + 7 – 0.5ﺇﺫﻥ A = 6.15 • ﻟﺤﺴﺎﺏ : B ﻫﻨﺎ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺴﺘﻌﻤﻠﺔ ﻟﻴﺴﺕ ﻜﻠﻬﺎ ﺃﻋﺩﺍﺩ ﻋﺸﺭﻴﺔ ﺇﺫﻥ ﻨﻘﻭﻡ ﺒﺘﻭﺤﻴﺩ ﺍﻟﻤﻘﺎﻤﺎﺕ ﺜﻡ ﻨﺠﻤﻊ B = 1×7×5 − 3×3×5 − 2×3×7 + 5×3×7 3×7×5 7×3×5 5×3×7 5×3×7 = 35 − 45 − 42 + 105 105 B = 53 105
• ﻟﺤﺴﺎﺏ : Cﺍﻷﻭﻟﻭﻴﺔ ﻹﻨﺠﺎﺯ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺩﺍﺨل ﺍﻷﻗﻭﺍﺱ )C =[(1.7)−(−1.8)]×(−2.5 )= (1.7+1.8)(−2.5 )C = (3.5)x(− 2.5 C = −8.75 • ﻟﺤﺴﺎﺏ : D ﻨﻅﺭﺍ ﻟﻁﻭل ﺍﻟﻌﻤﻠﻴﺎﺕ ,ﻨﺠﺯﺀ ﺍﻟﻌﻤل ﻜﻤﺎ ﻴﻠﻲ: D2 D1ﻭ ﻓﻨﺤﺴﺏ 1+ 1 ﺍﻟﻌﺩﺩ D2 ﻭ 1 − 1 ﻨﺴﻤﻲ D1ﺍﻟﻌﺩﺩ 1+ 1 1− 1 3 1 − 1 3 ﺜﻡ ﻨﺤﺴﺏ D • ﻟﺤﺴﺎﺏ : D1 1 = 1 ﻭ 1 − 1 =1− 3 ﻭ 1 = 3 ﻭ 1− 1 = 2 − 2 − 2 3 31 − 1 − 1 1 1 1 1 − 1 2 3 3 1 3 = −2 ﻭ ) D 1 = 1 − (− 2 D1 = 3 • ﻟﺤﺴﺎﺏ : D21 + 1 1 = 1+ 3 1 1 = 3 1 −ﻭ 1 = 4 + 3 4 3 4 3 3 1 ﻭ 1+
D2 = 7 4 =D D1 ﻤﻨﻪ : D2 D = 12 ﻭﺒﺎﻟﺘﺎﻟﻲ D ×= 3 4 = Dﻤﻨﻪ 3 7 7 7 4
• ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ : .1ﺃﺘﻤﻡ ﺍﻟﺠﺩﻭل ﺍﻟﺘﺎﻟﻲ ﺒﻭﻀﻊ ﺍﻟﻌﻼﻤﺔ xﻓﻲ ﺍﻟﺨﺎﻨﺎﺕ ﺍﻟﻤﻨﺎﺴﺒﺔ :5 − 0.7 105 − 14 − 5 1 1 3 7 3 2 xN xZ xD xQ xRﻤﺜﻼ ﻟﻘﺩ ﻭﻀﻌﺕ ﺍﻟﻌﻼﻤﺔ xﻓﻲ ﺠﻤﻴﻊ ﺍﻟﺨﺎﻨﺎﺕ ﺍﻟﻭﺍﻗﻌﺔ ﺘﺤﺕ 1ﻷﻥ 1∈ Nﻭ 1∈ Zﻭ 1∈ Dﻭ 1∈ Qﻭ 1∈R c, b, a .2ﺃﻋﺩﺍﺩ ﻏﻴﺭ ﻤﻌﺩﻭﻤﺔ ﻫل ﻴﻤﻜﻥ ﺇﺨﺘﺯﺍل ﺍﻟﻌﺒﺎﺭﺍﺕ C, B, Aﺍﻟﺘﺎﻟﻴﺔ ؟C = 5bc − 3ac , B = 3ab + 2b ,A = a + 2c ac b a .3ﺃﺤﺴﺏ ﺍﻷﻋﺩﺍﺩ E, D, C, B, Aﺍﻟﻤﻭﺍﻟﻴﺔ )ﺴﺘﻘﺩﻡ ﺍﻟﻨﺘﺎﺌﺞ ﻋﻠﻰ ﺸﻜل ﻜﺴﺭﻱ ﻏﻴﺭ ﻗﺎﺒل ﻟﻺﺨﺘﺯﺍل( A = 23.8 × 2.5 + 0.5 3.1×11.3 + 2.7 B = 5×212 +217 49 C= 232−1140+54−1140−1355−73
7 − 1 3 − 5 − 3 3 6 2 12 D = 2 − 5 1 × 10 + 3 3 3E = 2+ 1 + 1 1 1 2+ 1 6+ 1 2+ 2 6+ 6
• ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ:5 − 0.7 105 − 14 − 5 1 ﺕ : 01 3 7 3 2 1 x x x xN xx x xZ xD xxx xQ xR xxxxxxxxx ﺕ : 02 c,b,aﺃﻋﺩﺍﺩ ﺤﻘﻴﻘﻴﺔ ﻏﻴﺭ ﻤﻌﺩﻭﻤﺔ ﻟﻨﺒﺤﺙ ﺇﻥ ﻜﺎﻥ ﺒﺈﻤﻜﺎﻨﻨﺎ ﺍﻻﺨﺘﺯﺍل ﺒﺎﻟﻨﺴﺒﺔ ﻟﻜل ﻤﻥ ﺍﻟﻌﻴﺎﺭﺍﺕC, B, A : ﺃﻭﻻ :ﺍﻟﻌﺒﺎﺭﺓ : A A = a + 2c ﻟﺩﻴﻨﺎ : a ﺍﻟﺴﺒﺏ ﻓﻲ ﺃﻨﻨﺎ ﻻ ﻨﺴﺘﻁﻴﻊ ﺍﻻﺨﺘﺯﺍل ﻫﻭ ﻭﺠﻭﺩ ﺍﻟﻌﻤﻠﻴﺔ + ﻟﻭ ﻜﺎﻨﺕ ﺍﻟﻌﻤﻠﻴﺔ xﺒﺩﻻ ﻤﻥ +ﻟﺘﻤﻜﻨﺎ ﻤﻥ ﺍﻻﺨﺘﺯﺍل ﺜﺎﻨﻴﺎ :ﺍﻟﻌﺒﺎﺭﺓ : B B = 3ab + 2b = b(3a + )2 = 3a + 2 b b ﺜﺎﻨﻴﺎ :ﺍﻟﻌﺒﺎﺭﺓ : C ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ,ﻴﻤﻜﻨﻨﺎ ﺍﻻﺨﺘﺯﺍل ﺃﻴﻀﺎ ﻭﻫﺫﺍ ﻋﻠﻰ ﺍﻟﻨﺤﻭ ﺍﻟﺘﺎﻟﻲ: 5b 3a C = 5bc − 3ac = )c(5b − 3a = − ac ac a : 03 ﻟﻨﺤﺴﺏ ﺍﻷﻋﺩﺍﺩ E, D, C, B, A :ﻭﺴﻨﻘﺩﻡ ﺍﻟﻨﺘﺎﺌﺞ ﻋﻠﻰ ﺸﻜل ﻜﺒﻴﺭ ﻏﻴﺭ ﻗﺎﺒل ﻟﻼﺨﺘﺯﺍل
ﺃﻭﻻ ﺍﻟﻌﺩﺩ : A A = 23.8 × 2.5 + 0.5 = 59.5 + 0.5 = 60 3.1×11.3 + 2.7 35.03 + 2.7 37.73 ﺜﺎﻨﻴﺎ:ﺍﻟﻌﺩﺩ : B B = 5 ×15 + 217 = ×5 21× 21 + 7 × 31 = 7(5 × 21× 3 + )31 49 7×7 7×7 = 315 + 31 = 346 7 7 ﺜﺎﻟﺜﺎ :ﺍﻟﻌﺩﺩ : C C = 22 (− 14 + 4 ) − (1104 − 35 − ) 73 8 10 5 15= 22 × −14 + 8 − 14 × 21 − 35 ×14 − 3 × 30 = 11 × −3 − 294 − 490 − 90 8 10 ×3× 5× 2 7 4 5 210 = 11 (− )53 − 286 4 210 = 11 (− )53 + 143 = − 33 + 143 4 105 20 105 C = − 121 420 ﺭﺍﺒﻌﺎ:ﺍﻟﻌﺩﺩ : D 7 − 1 ( 3 − )5 − 3 7 − 1 × −7 − 3 3 6 2 12 3 6 2 12 D = = 5 10 3 10 2 − ( + 1 × 3 ) 2 − (5 × 10 × 3 ) 3 3 7 + 7 − 3 7 + 1 8 )13 −8 3 12 12 3 3 3 9 = = = × (− = 2−5 −3 E = 13 : ﺃﻥ ﺇﻟﻰ ﻨﺼل D ﺍﻟﻌﺩﺩ ﺤﺴﺎﺏ ﻓﻲ ﺍﻟﺴﺎﺒﻘﺔ ﺍﻟﻜﻴﻔﻴﺔ ﺒﻨﻔﺱ 3
ﺍﻷﻋﺩﺍﺩ ﺍﻷﻭﻟﻴﺔﻭﺍﻟﻘﻭﺍﺴﻡ ﻭﺍﻟﻤﻀﺎﻋﻔﺎﺕ ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ ) (N ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ : -ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﺃﻭﻟﻴﺔ ﻋﺩﺩ -ﺘﺤﻠﻴل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ -ﺤﺴﺎﺏ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻭﺍﻟﻤﻀﺎﻋﻑ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﺼﻐﺭ ﻟﻌﺩﺩﻴﻥﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ • ﺘﻌﺭﻴﻑ ﻭ ﻨﻅﺭﻴﺔ. • ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﺃﻭﻟﻴﺔ ﻋﺩﺩ ﺃﻜﺒﺭ ﻤﻥ .3• ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻌﺩﺩﻴﻥ ﻁﺒﻴﻌﻴﻴﻥﺍﻟﻤﻀﺎﻋﻑ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﺼﻐﺭ ﻟﻌﺩﺩﻴﻥ ﻁﺒﻴﻌﻴﻴﻥ • ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ • ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ •
• ﺘﻌﺭﻴﻑ ﻭ ﻨﻅﺭﻴﺔ: ﺘﻌﺭﻴﻑ: ﻴﻜﻭﻥ ﻋﺩﺩ ﻁﺒﻴﻌﻲ Pﺃﻭﻟﻴﺎ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ P≠1ﻭ Pﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ 1ﻭﻋﻠﻰ Pﻓﻘﻁ. ﻨﻅﺭﻴﺔ: ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ aﻏﻴﺭ ﺃﻭﻟﻲ ﻭﺃﻜﺒﺭ ﺘﻤﺎﻤﺎ ﻤﻥ 1ﻴﻘﺒل ﻋﻠﻰ ﺍﻷﻗل ﻗﺎﺴﻤﺎ ﺃﻭﻟﻴﺎ pﺒﺤﻴﺙ p2 ≤ a • ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﺃﻭﻟﻴﺔ ﻋﺩﺩ ﺃﻜﺒﺭ ﻤﻥ : 3 .ﻗﺎﻋﺩﺓ :ﻟﻴﻜﻥ aﻋﺩﺩﺍ ﻁﺒﻴﻌﻴﺎ ﺒﺤﻴﺙ a > 3ﺇﺫﺍ ﻜﺎﻥ aﻻ ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ ﺃﻱ ﻤﻥ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﻲ ﻤﺭﺘﺒﺘﻬﺎ ﺃﻗل ﻤﻥ ﺃﻭ ﺘﺴﺎﻭﻱ aﻓﺈﻥ aﺃﻭﻟﻲ. .ﺠﺩﻭل ﺍﻷﻋﺩﺍﺩ ﺍﻷﻭﻟﻴﺔ ﺍﻷﻗل ﻤﻥ : 100 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9 – 10 – 11 – 12 – 13 – 14 – 15 – 1617 – 18 - 19 – 20 – 21 – 22 – 23 – 24 – 25 – 26 – 27 – 28 – 29 – 30– – 31 – 32 – 33 – 34 – 35 – 36 – 37 – 38 – 39 – 40 – 41 – 42 – 4344 – 45 – 46 – 47 – 48 – 49 – 50 – 51 – 52 – 53 – 54 – 55 – 56 – 57– – 58 – 59 – 60 – 61 – 62 – 63 – 64 – 65 – 66 - 67- 68 – 69 – 7071 – 72 - 73 – 74 - 75 – 76 – 77 – 78 – 79 – 80 – 81 – 82 – 83 – 84– – 85 – 86 – 87 – 88 – 89 – 90 – 91 – 92 – 93 – 94 – 95 – 96 – 97 98 – 99 – 100
2ﺃﻭﻟﻲ ﻨﺸﻁﺏ ﻋﻠﻰ ﻤﻀﺎﻋﻔﺎﺕ 2ﺒﺩﺀﺍ ﻤﻥ 22 3ﺃﻭﻟﻲ ﻨﺸﻁﺏ ﻋﻠﻰ ﻤﻀﺎﻋﻔﺎﺕ 3ﺒﺩﺀﺍ ﻤﻥ 32 5ﺃﻭﻟﻲ ﻨﺸﻁﺏ ﻋﻠﻰ ﻤﻀﺎﻋﻔﺎﺕ 5ﺒﺩﺀﺍ ﻤﻥ 52 7ﺃﻭﻟﻲ ﻨﺸﻁﺏ ﻋﻠﻰ ﻤﻀﺎﻋﻔﺎﺕ 7ﺒﺩﺀﺍ ﻤﻥ 72ﻻ ﺩﺍﻋﻲ ﺃﻥ ﻨﻭﺍﺼل ,ﻷﻥ ﺍﻟﻌﺩﺩ ﺍﻷﻭﻟﻲ ﺍﻟﺫﻱ ﻴﺄﺘﻲ ﺒﻌﺩ 7ﻫﻭ 11ﻭ 100 < 112ﻤﻨﻪ ﺠﻤﻴﻊ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻐﻴﺭ ﻤﺸﻁﺏ ﻋﻠﻴﻬﺎ ﻫﻲ ﺃﻋﺩﺍﺩ ﺃﻭﻟﻴﺔ. ﻤﺜﺎﻻﻥ : • ﺍﻟﻤﺜﺎل ﺍﻷﻭل :ﻫل ﺍﻟﻌﺩﺩ 509ﺃﻭﻟﻲ ؟ﻟﻨﻘﺴﻡ 509ﻋﻠﻰ ﺍﻷﻋﺩﺍﺩ ﺍﻷﻭﻟﻴﺔ ﺍﻟﻤﺘﺴﺎﺒﻘﺔ ﺇﻟﻰ ﻏﺎﻴﺔ ﺃﻥ ﻨﺴﺘﻁﻴﻊ ﺍﻟﺤﻜﻡ 509ﻻ ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ 2 ﻭﻻ ﻋﻠﻰ 3ﻭﻻ ﻋﻠﻰ .5 509 = 7 x 72 + 5 509 = 11 x 46 + 3 509 = 17 x 29 + 16 509 = 19 x 26 + 15 509 = 23 x 22 + 3 ﻭﻟﺩﻴﻨﺎ 232 > 509ﻤﻨﻪ 509ﺃﻭﻟﻲ ﻤﻼﺤﻅﺔ :ﺤﺘﻰ ﻨﺘﺠﻨﺏ ﺤﺴﺎﺏ ﻤﺭﺒﻊ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﻓﻲ ﻜل ﻤﺭﺓ ,ﻻﺤﻅ ﺃﻥ ﺍﻟﻘﻭﺍﺴﻡ ﺘﺘﺼﺎﻋﺩ ﻭﺤﻭﺍﺼلﺍﻟﻘﺴﻤﺎﺕ ﺘﺘﻨﺎﻗﺹ ﻨﻜﺘﻔﻲ ﺒﺤﺴﺎﺏ ﻤﺭﺒﻊ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﻤﺘﺤﻥ ﻋﻨﺩﻤﺎ ﻨﺼل ﺇﻟﻰ ﻋﻤﻠﻴﺔ ﻗﺴﻤﺔ ﺤﺎﺼﻠﻬﺎ ﺃﻗل ﺘﻤﺎﻤﺎ ﻤﻥ ﻗﺎﺴﻤﻬﺎ. • ﺍﻟﻤﺜﺎل ﺍﻟﺜﺎﻨﻲ :ﻫل ﺍﻟﻌﺩﺩ 2561ﺃﻭﻟﻲ ؟ 2560ﻻ ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ 2ﻭﻻ ﻋﻠﻰ 3ﻭﻻ ﻋﻠﻰ 5 2561 = 7 x 365 + 6 2561 = 11 x 323 + 9 2561 = 13 x 197 + 0 ﻤﻨﻪ 2561ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ 13ﻭﺒﺎﻟﺘﺎﻟﻲ 2561ﻟﻴﺱ ﺃﻭﻟﻴﺎ
.ﺘﺤﻠﻴل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ ﺍﻟﻨﻅﺭﻴﺔ :ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ﻏﻴﺭ ﺃﻭﻟﻲ ﻭﺃﻜﺒﺭ ﺘﻤﺎﻤﺎ ﻤﻥ 1ﻴﻤﻜﻥ ﺘﺤﻠﻴﻠﻪ ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ ﻤﺜﺎل :ﺘﺤﻠﻴل 396ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ396 2198 299 333 311 11 • ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻌﺩﺩﻴﻥ ﻁﺒﻴﻌﻴﻴﻥ : ﺃ .ﺍﻟﻘﺎﻋﺩﺓ : ﻟﻠﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻌﺩﺩﻴﻥ ﻁﺒﻴﻌﻴﻴﻥ ﻜل ﻤﻨﻬﻤﺎ ﺃﻜﺒﺭ ﺘﻤﺎﻤﺎ ﻤﻥ: 1 .1ﻨﺤﻠل ﻜﻼ ﻤﻥ ﻫﺫﻴﻥ ﺍﻟﻌﺩﺩﻴﻥ ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ. .2ﻨﺤﺴﺏ ﺠﺩﺍﺀ ﺍﻟﻌﻭﺍﻤل ﺍﻷﻭﻟﻴﺔ ﺍﻟﻤﺸﺘﺭﻜﺔ ﺒﻴﻥ ﺍﻟﺘﺤﻠﻴﻠﻴﻥ ﺤﻴﺙ ﻴﺅﺨﺫ ﻜل ﻋﺎﻤل ﻤﻥ ﻫﺫﻩ ﺍﻟﻌﻭﺍﻤل ﻤﺭﺓ ﻭﺍﺤﺩﺓ ﻓﻘﻁ ﻭﺒﺄﺼﻐﺭ ﺃﺱ.ﻓﻲ ﺤﺎﻟﺔ ﻤﺎ ﻴﻜﻭﻥ ﺍﻟﺘﺤﻠﻴﻼﻥ ﻻ ﻴﺤﺘﻭﻴﺎﻥ ﻋﻠﻰ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ ﻤﺸﺘﺭﻜﺔ ﺒﻴﻨﻬﻤﺎ :ﻴﻜﻭﻥ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻠﻌﺩﻴﻥ ﺍﻟﻤﻘﺘﺭﺤﻴﻥ ﻫﻭ 1ﻭﻨﻘﻭل ﻋﻨﺩﺌﺫ ﻋﻥ ﻫﺫﻴﻥ ﺍﻟﻌﺩﺩﻴﻥ ﺃﻨﻬﻤﺎ ﺃﻭﻟﻴﺎﻥ ﻓﻴﻤﺎ ﺒﻴﻨﻬﻤﺎ. • ﻨﺭﻤﺯ ﺒﺎﻟﺭﻤﺯ ) (a, bﺇﻟﻰ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻠﻌﺩﺩﻴﻥ aﻭ b ﻤﺜﺎل 720 = 24 x 32 x 5 : 6300= 22 x 32 x 52 x 7 ﻭﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻠﻌﺩﻴﻥ 720ﻭ 6300ﻫﻭ 22 x 32 x 5 ﺃﻱ 180 ﺇﺫﺍ ﺭﻤﺯﻨﺎ ﺇﻟﻰ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﺒﺎﻟﺭﻤﺯ ) PGCDﻜﻤﺎ ﻫﻭ ﻤﻌﻤﻭل ﺒﻪ(.
PGCD (720 ,6300) = 180ﺏ .ﻁﺭﻴﻘﺔ ﺃﺨﺭﻯ ﻟﺤﺴﺎﺏ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻌﺩﺩﻴﻥ ﻁﺒﻴﻌﻴﻴﻥ : ﻟﺩﻴﻨﺎ6300 = 720 x 8 + 540 : 720 = 540 x 1 + 180 540 = 180 x 3 + 0 ﻟﺤﺴﺎﺏ )PGCD (720 ,6300 .1ﻗﻤﻨﺎ ﺒﺎﻟﻘﺴﻤﺔ ﺍﻹﻗﻠﻴﺩﻴﺔ ﻷﻜﺒﺭﻫﻤﺎ ﻋﻠﻰ ﺃﺼﻐﺭﻫﻤﺎ. .2ﻗﻤﻨﺎ ﺒﻤﻭﺍﺼﻠﺔ ﻋﻤﻠﻴﺎﺕ ﻗﺴﻤﺎﺕ ﺇﻗﻠﻴﺩﻴﺔ ﺒﺤﻴﺙ : ﺍﻟﻤﻘﺴﻭﻡ ﻟﻌﻤﻠﻴﺔ ﻫﻭ ﺍﻟﻘﺎﺴﻡ ﻟﺴﺎﺒﻘﺘﻬﺎ ﻭﺍﻟﻘﺎﺴﻡ ﻟﻌﻤﻠﻴﺔ ﻫﻭ ﺍﻟﺒﺎﻗﻲ ﻟﺴﺎﺒﻘﺘﻬﺎ ﻭﺘﻭﻗﻔﻨﺎ ﻋﻨﺩﻤﺎ ﺤﺼﻠﻨﺎ ﻋﻠﻰ ﺒﺎﻕ ﻤﻌﺩﻭﻡ )ﻻ ﻴﻤﻜﻥ ﺃﻥ ﺘﻘﺴﻡ ﻋﻠﻴﻪ (ﻓﻌﻨﺩﺌﺫ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻠﻌﺩﺩﻴﻥ ﻫﻭ ﺁﺨﺭ ﺒﺎﻕ ﻏﻴﺭ ﻤﻌﺩﻭﻡ )ﺃﻭ ﺁﺨﺭ ﻗﺎﺴﻡ( ﻤﻥ ﺒﻭﺍﻗﻲ )ﻗﻭﺍﺴﻡ( ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﻤﻨﺠﺯﺓ.ﻭﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﻤﺫﻜﻭﺭﺓ ﻫﻨﺎ ﺘﺴﻤﻰ \" ﻁﺭﻴﻘﺔ ﺤﺴﺎﺏ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻌﺩﺩﻴﻥ ﻁﺒﻴﻌﻴﻴﻥ ﺒﺎﺴﺘﻌﻤﺎل \" ﺨﻭﺍﺭﺯﻤﻴﺔ ﺇﻗﻠﻴﺩﺱ \" ﻁﺭﻴﻘﺔ ﻋﻤﻠﻴﺔ ﻟﺘﻨﻅﻴﻡ ﺍﻟﺤﺴﺎﺒﺎﺕ ﻟﻨﺤﺴﺏ ) PGCD (1800 ,1712ﺍﺴﺘﻌﻤﺎل ﺨﻭﺍﺭﺯﻤﻴﺔ ﺇﻗﻠﻴﺩﺱ ﺇﺫﻥ PGCD (1800 ,1712) = 8ﺤﺎﺼل 1 19 25ﻤﻘﺴﻭﻡ-ﻗﺎﺴﻡ 1800 1712 88 40 8 088 83240 8ﺒﺎﻗﻲ 0 ﺝ .ﺨﺎﺼﻴﺔ :ﻤﺠﻤﻭﻋﺔ ﺍﻟﻘﻭﺍﺴﻡ ﺍﻟﻤﺸﺘﺭﻜﺔ ﻟﻌﺩﺩﻴﻥ ﻁﺒﻴﻌﻴﻴﻥ ﻏﻴﺭ ﻤﻌﺩﻭﻤﻴﻥ ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﻗﻭﺍﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻬﺫﻴﻥ ﺍﻟﻌﺩﺩﻴﻥ.
• ﺍﻟﻤﻀﺎﻋﻑ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﺼﻐﺭ ﻟﻌﺩﺩﻴﻥ ﻁﺒﻴﻌﻴﻴﻥ :ﺃ .ﺍﻟﻘﺎﻋﺩﺓ :ﻟﻠﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻤﻀﺎﻋﻑ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﺼﻐﺭ ﻟﻌﺩﺩﻴﻥ ﻁﺒﻴﻌﻴﻴﻥ ﻜل ﻤﻨﻬﻤﺎ ﺃﻜﺒﺭ ﺘﻤﺎﻤﺎ ﻤﻥ :1 • .1ﻨﺤﻠل ﻜﻼ ﻤﻥ ﺍﻟﻌﺩﺩﻴﻥ ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ. .2ﻨﺤﺴﺏ ﺠﺩﺍﺀ ﻫﺫﻩ ﺍﻟﻌﻭﺍﻤل ﺒﺤﻴﺙ ﻴﻭﺨﺫ ﻜل ﻋﺎﻤل ﻤﺭﺓ ﻭﺍﺤﺩﺓ ﻓﻘﻁ ﻭﺒﺄﻜﺒﺭ ﺃﺱ. ﻨﺭﻤﺯ ﺒﺎﻟﺭﻤﺯ ) (a, bﺇﻟﻰ ﺍﻟﻤﻀﺎﻋﻑ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﺼﻐﺭ ﻟﻠﻌﺩﺩﻴﻥ aﻭ .b ﻤﺜﺎل : 792 = 23 x 32 x 11 2100 = 22 x 3 x 52 x 7PPCM (792 ,2100) = 23 x 32 x 52 x 11 = 138600 ﺏ .ﺨﺎﺼﻴﺔ :ﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﻀﺎﻋﻔﺎﺕ ﺍﻟﻤﺸﺘﺭﻜﺔ ﻟﻌﺩﺩﻴﻥ ﻁﺒﻴﻌﻴﻴﻥ ﻏﻴﺭ ﻤﻌﺩﻭﻤﻴﻥ ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﻤﻀﺎﻋﻔﺎﺕ ﺍﻟﻤﻀﺎﻋﻑ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﺼﻐﺭ ﻟﻬﺫﻴﻥ ﺍﻟﻌﺩﺩﻴﻥ.
• ﺕﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ : .1ﻗل )ﻤﻊ ﺍﻟﺘﻌﻠﻴل( ﻋﻥ ﻜل ﻤﻥ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﺎﻟﻴﺔ ﺇﻥ ﻜﺎﻥ ﺃﻭﻟﻴﺎ : .1277 ,119 ,277 ,407 ,667 ,373 .2ﺤﻠل ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ ﻜﻼ ﻤﻥ ﺍﻷﻋﺩﺍﺩ 16335 ,2730 ,1728 ,312 ,180 .3ﺤﻠل ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ ﻜﻼ ﻤﻥ ﺍﻷﻋﺩﺍﺩ c, b,aﺍﻟﻤﻭﺍﻟﻴﺔ a = 16x10x15x18 b=8x36x20x18x4a5 c =17x48x25x32x63 .4ﺃﺤﺴﺏ ﺍﻟﻘﺎﺴﻤﺎﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻠﻌﺩﺩﻴﻥ aﻭ bﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ : a = 630 .1ﻭ b = 720 a = 810 .2ﻭb = 590 a = 2730 .3ﻭb = 2310 .5ﺃﺤﺴﺏ ﺍﻟﻤﻀﺎﻋﻑ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﺼﻐﺭ ﻟﻠﻌﺩﺩﻴﻥ aﻭ bﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ : a = 675 .1ﻭ b = 1800 a = 640 .2ﻭb = 910 a = 2910 .3ﻭb = 3465 n .6ﻋﺩﺩ ﺘﻼﻤﻴﺫ ﻤﺘﺭﺸﺤﻴﻥ ﺇﻟﻰ ﺍﻤﺘﺤﺎﻥ ,ﺒﺤﻴﺙ nﺃﻗل ﻤﻥ 700ﻭ nﺃﻜﺒﺭ ﻤﻥ 600ﻴﺭﺍﺩ ﺘﻔﻭﻴﺞﻫﺅﻻﺀ ﺍﻟﺘﻼﻤﻴﺫ ﻓﻠﻭﺤﻅ ﻤﺎ ﻴﻠﻲ ﺇﺫﺍ ﻜﻭﻨﺕ ﺃﻓﻭﺍﺝ ﺘﻀﻡ 20ﺘﻠﻤﻴﺫﺍ ﺃﻭ ﺃﻓﻭﺍﺝ ﺘﻀﻡ 24ﺘﻠﻤﻴﺫﺍ ﻓﻲ ﻜل ﻤﺭﺓ ﻴﺒﻘﻰ 9ﺘﻼﻤﻴﺫ ﻏﻴﺭ ﻤﻔﻭﺠﻴﻥ. ﻤﺎ ﻫﻭ ﻋﺩﺩ ﻫﺅﻻﺀ ﺍﻟﺘﻼﻤﻴﺫ ؟
• ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ: ﺘﻤﺭﻴﻥ : 01 ﺘﺤﺩﻴﺩ ﺍﻟﻌﺩﺩ ﺍﻷﻭﻟﻲ ﻤﻥ ﺒﻴﻥ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻁﺒﻴﻌﻴﺔ ﺍﻟﻤﻌﻁﺎﺓ :• 373ﻻ ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ ﺃﻱ ﻋﺩﺩ ﺃﻭﻟﻲ ﻤﻥ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﺎﻟﻴﺔ : 23 ،19 ،17 ،13 ،11 ،7 ،3 ،2 ﻭﻟﻜﻥ 373 < 232 :ﻭﻤﻨﻪ 373 :ﺃﻭﻟﻲ. • 667ﻟﻴﺱ ﺃﻭﻟﻴﺎ ﻷﻥ29 × 23 = 667 : ﺃﻱ 667ﻴﻘﺒل ﺃﻜﺜﺭ ﻤﻥ ﻗﺎﺴﻤﻴﻥ. ﻭﺒﻨﻔﺱ ﺍﻟﻜﻴﻔﻴﺔ ﻴﻜﻭﻥ ﺍﻟﺘﻌﺎﻤل ﻤﻊ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻁﺒﻴﻌﻴﺔ ﺍﻟﻤﺘﺒﻘﻴﺔ. ﺘﻤﺭﻴﻥ : 02 ﻟﻨﺤﻠل ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ ﻜﻼ ﻤﻥ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﻌﻁﺎﺓ. • 5 × 32 × 22 = 180 • 13× 3× 23 = 312 • 33 × 26 = 1728 • 13 × 7 × 5 × 3× 2 = 2730 • 112 × 5 × 33 = 16335 ﺘﻤﺭﻴﻥ : 03 ﻟﻨﺤﻠل ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ ﻜﻼ ﻤﻥ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﻌﻁﺎﺓ • a = 26 × 33 × 52 • b = 211 × 34 × 52 • c = 29 × 33 × 52 × 7 ×17
ﺘﻤﺭﻴﻥ : 04 ﻟﻨﺤﺴﺏ ﺍﻟﻘﺎﺴﻡ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﻜﺒﺭ ﻟﻠﻌﺩﺩﻴﻥ aﻭ bﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﻌﻁﺎﺓ : a = 630 .1ﻭ b = 720 ﻟﺩﻴﻨﺎ a = 2 × 32 × 5 × 7 :ﻭ b = 24 × 32 × 5 ﻭﻤﻨﻪ P.G.C.D(a,b) = 2 × 32 × 5 = 90 : a = 810 .2ﻭ b = 590ﺤﺎﺼل ﻟﻨﺤﺴﺏ ) PGCD(810,590ﺒﺎﺴﺘﻌﻤﺎل ﺨﻭﺍﺭﺯﻤﻴﺔ ﺇﻗﻠﻴﺩﺱﻤﻘﺴﻭﻡ- 12127 ﻗﺎﺴﻡ ﺒﺎﻗﻲ 810 590 220 150 70 10 220 150 70 0 .3ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﻁﺭﻴﻘﺘﻴﻥ ﺍﻟﺴﺎﺒﻘﺘﻴﻥ ﻨﺠﺩ ﺃﻥ : PGCD(2730,2310) = 210 ﺘﻤﺭﻴﻥ : 05 ﻟﻨﺤﺴﺏ ﺍﻟﻤﻀﺎﻋﻑ ﺍﻟﻤﺸﺘﺭﻙ ﺍﻷﺼﻐﺭ ﻟﻠﻌﺩﺩﻴﻥ aﻭ bﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﻌﻁﺎﺓ : a = 675 .1ﻭ b = 1800 ﻟﺩﻴﻨﺎ a = 33 × 52 :ﻭ b = 23 × 32 × 52 ﻭﻤﻨﻪ PPCM (720,630) = 23 × 33 × 52 : = 5400 a =640 .2ﻭ b = 910 ﻟﺩﻴﻨﺎ b = 2 × 5 × 7 ×13 :ﻭ b = 27 × 5 ﻭﻤﻨﻪ PPCM (640,910) = 27 × 5 × 7 ×13 : = 58240 a = 2910 .3ﻭ b = 3465 ﻟﺩﻴﻨﺎ a = 2 × 3 × 5 × 97 :ﻭ b = 32 × 5 × 7 ×11 ﻭﻤﻨﻪPPCM (2910,3465) = 2 × 32 × 5 × 7 ×11 : = 672210 < 97
ﺘﻤﺭﻴﻥ : 06 ﺘﻌﻴﻴﻥ nﺍﻟﺠﻤﻠﺔ \" ﺇﺫﺍ ﻜﻭﻨﺎ ﺃﻓﻭﺍﺝ ﺘﻀﻡ 20ﺘﻠﻤﻴﺫ ﻴﺒﻘﻰ ﻏﻴﺭ ﻤﻔﻭﺠﻴﻥ \" ﻴﻤﻜﻨﻨﺎ ﺼﻴﺎﻏﺘﻬﺎ ﻋﻠﻰ ﺍﻟﻨﺤﻭ:) n = a × 20 + 9ﺤﻴﺙ a :ﻋﺩﺩ ﺍﻷﻓﻭﺍﺝ(ﻭﺍﻟﺠﻤﻠﺔ \" ﺇﺫﺍ ﻜﻭﻨﺎ ﺃﻓﻭﺍﺠﺎ ﺘﻀﻡ 24ﺘﻠﻤﻴﺫ ﻴﺒﻘﻰ 9ﺘﻼﻤﻴﺫ ﻏﻴﺭ ﻤﻔﻭﺠﻴﻥ \" ﻴﻤﻜﻨﻨﺎ ﺼﻴﺎﻏﺘﻬﺎ ﻋﻠﻰ ﺍﻟﻨﺤﻭ :)ﺤﻴﺙ b :ﻋﺩﺩ ﺍﻷﻓﻭﺍﺝ( n = b × 24 + 9n − 9 = a × 20ﻭﻋﻠﻴﻪ ﻴﺼﺒﺢ ﻟﺩﻴﻨﺎ n − 9 = b × 24 :ﻭﻋﻠﻴﻪ n − 9ﻤﻀﺎﻋﻑ ﻟـ 20ﻭﻤﻀﺎﻋﻑ ﻟـ .24ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻨﻪ ﻤﻀﺎﻋﻑ ﻤﺸﺘﺭﻙ ﻓﻬﻭ ﺇﺫﻥ) (ﻤﻀﺎﻋﻑ ﻟﻠﻌﺩﺩ PPCM (24,20) :ﺃﻱ (n − 9) :ﻤﻀﺎﻋﻑ ﻟـ120 :ﻭﺒﺎﻟﺘﺎﻟﻲ n − 9 = 120 × k : k ∈ N : ﺃﻱ n = 120k + 9 :ﻤﻊ ﺍﻟﻌﻠﻡ ﺃﻥ 600 ≤ n ≤ 700 :ﻭﻤﻨﻪ k = 5 :ﻭﻤﻨﻪn = 609 :
ﺍﻟﻘﻭﻯ ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ: -ﺍﻟﺘﻤﻜﻥ ﻤﻥ ﺘﻌﺭﻴﻑ ﺍﻟﻘﻭﻯ ﻭﺨﻭﺍﺼﻬﺎ -ﺍﻟﺘﺤﻜﻡ ﻓﻲ ﺍﻟﺤﺴﺎﺒﺎﺕ ﻋﻠﻰ ﺍﻟﻘﻭﻯﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ • ﺘﻌﺭﻴﻑ • ﺨﻭﺍﺹ ﺍﻟﻘﻭﻯ • ﺠﺩﺍﺀﺍﺕ ﺸﻬﻴﺭﺓ • ﻗﻭﻯ ﺍﻟﻌﺩﺩ ﻋﺸﺭﺓ • ﺍﺴﺘﻌﻤﺎل ﺍﻵﻟﺔ ﺍﻟﺤﺎﺴﺒﺔ ﻟﺤﺴﺎﺏ ﻗﻭﻯ • ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ • ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ
• ﺘﻌﺭﻴﻑ :ﺇﺫﺍ ﻜﺎﻥ aﻋﻨﺼﺭﺍ ﻤﻥ *) Rﺃﻱ a ∈ Rﻭ ( a ≠ 0ﺍﻟﻘﻭﻯ ﺫﺍﺕ ﺍﻷﺱ ﻓﻲ Zﻟﻠﻌﺩﺩ aﻤﻌﺭﻓﺔ ﻜﻤﺎ • ﻴﻠﻲ : a = a1 a0 =1ﻭﻭﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﻁﺒﻴﻌﻲ .n An = axax…xaﺇﺫﺍ ﻜﺎﻥ n ≥ 2 Nﻋﺎﻤﻼ a −n = 1 an • ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﻁﺒﻴﻌﻲ ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻡ 0n = 0: n • ﺨﻭﺍﺹ ﺍﻟﻘﻭﻯ :ﻤﻥ ﺃﺠل ﻋﺩﺩﻴﻥ ﺤﻘﻴﻘﻴﻴﻥ ﻏﻴﺭ ﻤﻌﺩﻭﻤﻴﻥ aﻭ bﻭﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩﻴﻥ ﺼﺤﻴﺤﻴﻥ ﻨﺴﺒﻴﻴﻥ ﻭ pﻟﺩﻴﻨﺎ : ap ×aq =ap+qﻭ a×b p = ap ×bpﻭ ( ) ( )ap q = apxq a p = ap ﻭ b bpap = a p−q a−pﻭ = 1 ﻭaq an
: • ﺠﺩﺍﺀﺍﺕ ﺸﻬﻴﺭﺓ b ﻭa ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﺤﻘﻴﻘﻴﺎﻥ(a+b) x(a-b)= a2-b2 (a-b)2=a2+ab+b2 (a+b)2=a2+ab + b2(a+-b)3 = a3+3a2b+3ab2+b3 (a-b)3=a3-3a2b+3ab2-b3a3+b3=(a+b)(a2-ab+b2) a3-b3=(a-b)(a2+ab+b2) \" • ﺍﻟﺒﺭﻫﺎﻥ ﻋﻠﻰ ﺍﻟﻨﺘﺎﺌﺞ \" ﺍﻟﺠﺩﻴﺩﺓ (a + b)3 = (a + b)2( a + b) = (a2 + 2ab + b2)(a + b) = a3 + 2a2b + ab2 + ba2 + 2ab2 + b3 = a3 + 3a2b + 3ab2 + b3 (a-b)3 = (a-b)2 (a-b) = (a2 - 2ab + b2)(a - b) = a3 - 2a2b + ab2 - a2b + 2ab2 -b3 = a3 - 3a2b + 3ab2 - b3 (a+b)(a2-ab+b2) = a3-a2b+ab2+ba2-ab2+b3 = a3 + b3 (a - b)(a2 + ab + b2) = a3 + a2b + ab2 - a2b - ab2 - b3 = a3 - b3 : • ﺃﻤﺜﻠﺔ x ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ:ﻟﻠﻨﺸﺭ (x + 1)3 = x3 + 3x21 + 3x12 + 13 (x + 1)3 = x3 +3x2 + 3x + 1 (2x - 3)3 = (2x)3 - 3(2x)23 + 3.2x.32 - 33 = 23x3-3.2.x2.3+27.2x-27 (2x - 3)3 = 8x3 -36x2 -54x - 27
ﻟﻠﺘﺤﻠﻴل :ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ x X3 – 1 = x3 - 13 )= (x - 1)(x2 + 1.x + 12 )= (x - 1)(x2 + x + 1 x3 + 8 = x3 + 23 )= (x + 2)(x2 - 2x + 22 )= (x + 2)(x2 - 2x + 4 • ﻗﻭﻯ ﺍﻟﻌﺩﺩ ﻋﺸﺭﺓ : ﻟﺩﻴﻨﺎ100 =1 :ﻭ 101=10ﻭ 102=100ﻭ103 = 1000 ﻭ 10 1-= 0.1ﻭ 10 2-= 0.01ﻭ 10 3 -= 0.001 ﻭﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﻁﺒﻴﻌﻲ ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻡﻭ10-n = 0,00….01 10n =100....0ﻋﺩﺩ ﺍﻷﺭﻗﺎﻡ ﺒﻌﺩ ﺍﻟﻔﺎﺼﻠﺔ ﻫﻭn ﻋﺩﺩ ﺍﻷﺼﻔﺎﺭ ﻫﻭnﻤﻼﺤﻅﺔ :ﺍﺴﺘﻌﻤﺎل ﺍﻟﻘﻭﻯ ﻋﺸﺭﺓ ﺘﻤﻜﻥ ﻤﻥ ﺘﺒﺴﻴﻁ ﺒﻌﺽ ﺍﻟﺤﺴﺎﺒﺎﺕ ﻋﻠﻰ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻌﺸﺭﻴﺔ=A 0 .000025 × 0 .0003 ﻤﺜﺎل :ﻟﻨﺤﺴﺏ ﺍﻟﻌﺩﺩ Aﺤﻴﺙ 0 .00075 A = 25×10−6 × 3×10−4 : ﻟﺩﻴﻨﺎ 75×10−5 A = 25 × 3 ×10 −10 75 ×10 −5
A = 75 × 10 − 10 75 × 10 − 5 A = 10 − 10 10 − 10 )A =10−10−(−5 A = 10 −5 A = 0.00001 • ﺍﺴﺘﻌﻤﺎل ﺍﻵﻟﺔ ﺍﻟﺤﺎﺴﺒﺔ ﻟﺤﺴﺎﺏ ﻗﻭﻯ: 32ﺇﺫﻥ 325=3355.4432 ﻤﺜﺎل :ﻟﺤﺴﺎﺏ 5 = : 325 ﺇﺫﻥ(23 -)3 = -12167 : ﻟﺤﺴﺎﺏ < (.) 23 > xy 3 = :(23-)3xy ﻟﺤﺴﺎﺏ :(25-173)-212 :(252-173)-212=-4729ﻨﺠﺩ< 25 xy 2-17 xy = 3> -21 xy 2
• ﺕﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ : .1ﺃﺤﺴﺏ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﺎﻟﻴﺔ : (-8)-3 ,8-8 ,(-3)-8 ,3-8 ,(-8)3 ,(-3) 8 ,83 ,38 .2ﺒﺴﻁ ﻋﺒﺎﺭﺓ ﻜل ﻤﻥ ﺍﻷﻋﺩﺍﺩ C. B. A. )ﻓﻜﺭ ﻓﻲ ﺘﺤﻠﻴل ﺍﻷﻋﺩﺍﺩ ﺍﻟﻁﺒﻴﻌﻴﺔ ﺇﻟﻰ ﺠﺩﺍﺀ ﻋﻭﺍﻤل ﺃﻭﻟﻴﺔ( − 2112 × − 45 −7 517 ×35−10 = ( ) ( )B = ,A 35 9 × 20 6 14 9 × 25 8 C = 1059 ×(− 60)6 (− 70)9 ×98 0.1 + 1 − 0 .02 .3ﺃﺤﺴﺏ ﺍﻷﻋﺩﺍﺩ C, B, Aﺍﻟﺘﺎﻟﻴﺔ 2000 0 .01 B = , A = 1000 −4 × (0.001)5 ×100 −1 × 2 × 0.5 C = (2.7 )3 × (0.15 )−4 × 0.05 810000 z, y, x .4ﺃﻋﺩﺍﺩ ﺤﻘﻴﻘﺔ ﻏﻴﺭ ﻤﻌﺩﻭﻤﺔ ﻨﻌﺘﺒﺭ ﺍﻟﻌﺩﺩﻴﻥ aﻭ bﺤﻴﺙ :( ) ( ) ( ) ( )b= x7 × 1 × (xy)3 × −5 × 1 =,a xy 4 × xy−1 −2 × xyz 3 x −5 x14 z 4 −3 x5 y7 z−1 ﺒﺴﻁ ﻋﺒﺎﺭﺓ ﻜل ﻤﻥ aﻭ b .1 .2 z = 1 = ,y=2 , x 1 ﺃﺤﺴﺏ ﻗﻴﻤﺔ ﻜل ﻤﻥ aﻭ bﻤﻥ ﺃﺠل 3 27
• ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ: ﺘﻤﺭﻴﻥ : 01ﺤﺴﺎﺏ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﻌﻁﺎﺓ ، 38 = 6561 ، 83 = 512 ، (− 3)8 = 6561 ، (− )8 −3 = −0.001953125 ، (− 8)3 = −512 ، 58 = 0.00015241579003 (− )3 −8 = 3−8 ﺘﻤﺭﻴﻥ : 02ﺘﺒﺴﻴﻁ ﻜل ﻤﻥ ﺍﻷﻋﺩﺍﺩ C, B, A ( ( ) )A 6 359 × 20 6 (5 × 7)9 × 22 × 5 59 × 79 × 212 × 56 = 149 × 258 = (2 × 7)9 × 52 8 = 29 × 79 × 516 :A ﺍﻟﻌﺩﺩ .1 =A 212 × 515 × 79 = 23 × 5−1 = 8 = 1.6 29 × 516 × 79 5 .2ﺍﻟﻌﺩﺩ : B ( )B −7 (− 21)12 × (− 45)−7 (− 3× 7)12 × − 32 × 5 = = ( )517 × 5 × 7 −10 517 × 35−10 = 312 × 712 × 510 × 710 512 × 314 × 57 .3ﺍﻟﻌﺩﺩ : C( ( ) )C 6 1059 × (− 60)6 (3× 5 × 7)9 × 22 × 3× 5 = (− 70)9 × 98 = − (2 × 5 × 7)9 × 32 8= 39 × 59 × 79 × 212 × 36 × 56 = − 23 × 56 − 29 × 59 × 79 × 316 3= − 125000 3
ﺘﻤﺭﻴﻥ :03ﺤﺴﺎﺏ ﻜل ﻤﻥ C, B, A .1ﺍﻟﻌﺩﺩ : B 0.1 + 1 − 0.02 0.1 + 0.0005 − 0.02 2000 0.01 B = = 2 × 0.5 × 0.01 = 0.0805 = 8.05 ×10−2 = 8.05 0.01 1 × 10 −2 .2ﺍﻟﻌﺩﺩ :A ( ) ( ) ( )( )A = 1000−4 × 0.001 5 ×100−1 = 103 −4 × 10−3 5 × 102 −1 ﺃﻱ A = 10 −12 × 10 −15 × 10 −2 = 10 −29 .3ﺍﻟﻌﺩﺩ :C( ) ( ) ( )C × −4= (2.7)3 × (0.15)−4 × )(0.05 = 27 ×10−1 3 × 15 ×10−2 5 ×10−2 81 × 10 4 810000 = 273 ×10−3 ×15−4 ×108 × 5 ×10−2 = 34 39 × 5 ×10−1 81 × 10 4 × 34 × 54 ×104 = 53 3 = 0.0024 ×10 ﺕ : 04 Z, y, xﺃﻋﺩﺍﺩﺍ ﺤﻘﻴﻘﻴﺔ ﻏﻴﺭ ﻤﻌﺩﻭﻤﺔ .1ﺘﺒﺴﻴﻁ ﻋﺒﺎﺭﺓ ﻜل ﻤﻥ aﻭ b = x 4 y 4 x −2 y 2 x3 y 3 z 3 a( )، = (xy)4 × xy −1 −2 × (xyz)3 ﻟﺩﻴﻨﺎ: x5 y 7 z −1 x5 × y7 × z −1 = y2z4 = x5 y9z3 x5 y 7 z −1
b = x7 × 1 × (xy)3 × 3−5 × 1 :ﻭﻟﺩﻴﻨﺎ x −5 x14 ( yz )−3 = x7 ×x5 × x3 × y3 ×z−5 × 1 × y3 ×z3 x14 = x15 × y6 × z3 = x× y6 x14 × z5 z2z = 1 ، y = 2 ، x = 1 : ﻤﻥ ﺃﺠلbﻭ a ﺤﺴﺎﺏ ﻗﻴﻤﺔ ﻜل ﻤﻥ .1 3 27 a= y2z4 × 1 4 4 3 81 = 22 = : ﻟﺩﻴﻨﺎ x× y6 1 × 26 64 32 27 3 b = = 2 = : ﻭﻟﺩﻴﻨﺎ 1 3
ﺍﻟﺠﺫﻭﺭ ﺍﻟﺘﺭﺒﻴﻌﻴﺔ ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ: -ﺍﻟﺘﻤﻜﻥ ﻤﻥ ﺘﻌﺭﻴﻑ ﺍﻟﺠﺫﺭ ﺍﻟﺘﺭﺒﻴﻌﻲ ﻟﻌﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﻭﺨﻭﺍﺹ ﺍﻟﺠﺫﻭﺭ ﺍﻟﺘﺭﺒﻴﻌﻴﺔ -ﺍﻟﺘﺤﻜﻡ ﻓﻲ ﺍﻟﺤﺴﺎﺒﺎﺕ ﻋﻠﻰ ﺍﻟﺠﺫﻭﺭ ﺍﻟﺘﺭﺒﻴﻌﻴﺔ ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ • ﺘﻌﺭﻴﻑ ,ﺘﺭﻤﻴﺯ. • ﺨﻭﺍﺹ ﺍﻟﺠﺫﻭﺭ ﺍﻟﺘﺭﺒﻴﻌﻴﺔ. • ﺍﺴﺘﻌﻤﺎل ﺍﻵﻟﺔ ﺍﻟﺤﺎﺴﺒﺔ ﻟﺤﺴﺎﺏ ﻗﻭﻯ • ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ • ﺍﻟﺤﻠﻭل
• ﺘﻌﺭﻴﻑ ,ﺘﺭﻤﻴﺯ:ﺍﻟﺠﺫﺭ ﺍﻟﺘﺭﺒﻴﻌﻲ ﺍﻟﻤﻭﺠﺏ ﻟﻌﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ aﻫﻭ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ﺍﻟﻤﻭﺠﺏ ,ﺍﻟﻤﺭﻤﺯ ﺇﻟﻴﻪ ﺒﺎﻟﺭﻤﺯ ,ﺍﻟﺫﻱ ﻤﺭﺒﻌﻪ ﻴﺴﺎﻭﻱ .a aﻭ b≥0 a ≥ 0, ﺃﻱ ﻤﻥ ﺃﺠل aﻭ bﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ﺒﺤﻴﺙ: b2 =aﻴﻌﻨﻲ b = a ﺃﻤﺜﻠﺔ 81 = 9 , 4 = 2 : (− 3)2 = 9 = 3 ﻭﻟﻜﻥ 9 ≠ −3ﺭﻏﻡ ﺃﻥ ( )9= −3 2 ﻋﺩﺩ ﻤﻭﺠﺏ ﺤﺴﺏ ﺍﻟﺘﻌﺭﻴﻑ ‼!9 ﻭﻟﻜﻥ
• ﺨﻭﺍﺹ ﺍﻟﺠﺫﻭﺭ ﺍﻟﺘﺭﺒﻴﻌﻴﺔ : • ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ﺍﻟﻤﻭﺠﺏ a a2 = aﻭ ( )a 2 = a • ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﺤﻘﻴﻘﻴﺎﻥ ﺍﻟﻤﻭﺠﺒﺎﻥ aﻭ b )(a×b = a)×( b• ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ﺍﻟﻤﻭﺠﺏ ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻕ ﺍﻟﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ b a = a b b• ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻕ ﺍﻟﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ aﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺼﺤﻴﺢ ﺍﻟﻨﺴﺒﻲ :n ( )an = a n ﺃﻤﺜﻠﺔ : 32 16× 2 =( 16)× 2 =4 275 = 3× 2564 64
= 3 × 25 64= 3 × 25 64= 53 8720 = 24 ×32 ×5= 24 × 32 × 5( )= 22 2 ×3× 5= 22 ×3× 5 = 12 5 16+ 9 = 4+3 =7 16+ 9 = 25 =5 ﻭﻟﻨﺎ 5 ≠ 7ﺇﺫﻥ ﺍﻟﺠﺫﺭ ﺍﻟﺘﺭﺒﻴﻌﻲ ﻟﻤﺠﻤﻭﻉ ﻋﺩﺩﻴﻥ ﻤﻭﺠﺒﻴﻥ ﻻ ﻴﺴﺎﻭﻱ ﺒﺎﻟﻀﺭﻭﺭﺓ ﻤﺠﻤﻭﻉ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﺘﺭﺒﻴﻌﻴﻴﻥ ﻟﻬﺫﻴﻥ ﺍﻟﻌﺩﺩﻴﻥ.
• ﺍﺴﺘﻌﻤﺎل ﺍﻵﻟﺔ ﺍﻟﺤﺎﺴﺒﺔ:ﺘﻤﻜﻥ ﻤﻥ ﺤﺴﺎﺏ ﺍﻟﺠﺫﻭﺭ ﺍﻟﺘﺭﺒﻴﻌﻴﺔ ﻓﻲ ﺒﻌﺽ ﺍﻟﺤﺴﺎﺒﺎﺕ ﻨﻀﻐﻁ ﻋﻠﻰ ﺍﻟﻠﻤﺴﺔﻗﺒل ﺍﻟﻌﺩﺩ ﻭﻓﻲ ﺍﻟﺒﻌﺽ ﺍﻵﺨﺭ ﺒﻌﺩ ﺍﻟﻌﺩﺩ ‼ ﻭﻟﻜﻥ ﺇﺫﺍ ﻜﺎﻥ aﻋﺩﺩﺍ ﻤﻭﺠﺒﺎ ﻟﻴﺱ ﻤﺭﺒﻊ ﻋﺩﺩ ﻋﺸﺭﻱ ﻓﺎﻟﺩﺍﻟﺔ ﺘﻌﻁﻲ ﻗﻴﻤﺔ ﺘﻘﺭﻴﺒﻴﺔ ﻓﻘﻁ ﻟﻠﻌﺩﺩ . aﻤﺜﺎل :ﻟﺤﺴﺎﺏ 17424 = :ﻓﻨﺠﺩ1742417424 = 132
• ﺘﻤﺎﺭﻴﻥ ﻭ ﻤﺸﻜﻼﺕ : .1ﺃﺤﺴﺏ ﺍﻷﻋﺩﺍﺩ H, G,F,E, D, C, B, Aﺍﻟﺘﺎﻟﻴﺔ := ,C 0.27 × 0.3 , B = 0.00012 , A= 49×36×25 14.4( ), E = 9 +18 5 2 ( ) ( ), D= 5+ 7 2 + 5− 7 2 ( ) ( )H = 2+ 2 − 2− ( )2−2 2 2+ 2− 5 ,G = 57× 79 2 2 35 5 ﺤﻴﺙ a ∈Rﻭ q ∈Q a .2ﺃﻜﺘﺏ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﺎﻟﻴﺔ ﻋﻠﻰ ﺍﻟﺸﻜل q 19 5 + 1 − 17 5 2 3 + 2 2 , 3 − 2 ,1+ 3 , 3 − 5 .3ﻟﻴﻜﻥ aﻭ bﺍﻟﻌﺩﺩﻴﻥ ﺤﻴﺙ : a = 11+ 3 7ﻭ b = 11−3 7 .1ﺃﺤﺴﺏ a2+b2ﻭ a x b .2ﺍﺴﺘﻨﺘﺞ (a+b)2ﻭ (a - b)2 .3ﺍﺴﺘﻨﺘﺞ ) (a+bﻭ )(a - b (1.4ﺃﺤﺴﺏ ( )2 + 5 2 (2ﺃﺤﺴﺏ 9 + 4 5 × 5 − 2 × 5 + 2 = ( ) ( )3 − 7 = 7 + 2 7− 1 = 1 6 .5ﺃﺜﺒﺕ ﺃﻨﻪ 7 + 1 2 7 −2 33+ 7
• ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ: ﺕ : 01ﺤﺴﺎﺏ ﻜل ﻤﻥ ﺍﻷﻋﺩﺍﺩG, F, E, D, C, B, A :ﻟﺩﻴﻨﺎB = 0.000121 = 10−6 ×121 : .1 ( )= 10−3 2 × (11)2 = 10−3 ×11 = 0.011 .2ﻭﻟﺩﻴﻨﺎ A = 49 × 36 × 25 = 72 × 62 × 52 = 72 × 62 × 52 = 7 × 6× 5 = 210 .3ﻟﺩﻴﻨﺎC = 0.4 × 0.3 = 32 × 3×10−2 × 10−1 × 3 : 14.4 10−1 ×122× = 3×10−1 × 3 × 10−1 3 = 9 ×10−1 = 3 ×10 −1 = 0.075 10−1 × 122 12 4 .4ﻭﻟﺩﻴﻨﺎ( ) ( )D = 5 + 7 2 + 5 − 7 2 : = 5 + 2 5 7 + 7 + 5 − 2 5 × 7 + 7 = 24 .5ﻭﻟﺩﻴﻨﺎ( )E = 9 +18 5 2 = 9 +18 5 : × 5 2 3 × 5 × 7 2 4 35 2 2 × 35= ( () () ) ( ) ( ) ( )F5779 7 × 5 = .6ﻭﻟﺩﻴﻨﺎ: 35= 53 ×5 × × 74 7 = × 53 × 5 × 74 7 = 51 × 72 = 245 352 × 35 × 52 ×72 × 5 7 1
G = 2+ 2 − 2− 2− 5 ﻭﻟﺩﻴﻨﺎ: .1 2−2 2 2+ 2 2 = (2 + ) (2 2 − 2 − 2 2)(2 − 2)− 5 )(2 − 2 2)(2 + 2 2 = 4+4 2+2−4+2 2+4 2−4− 5 4+2 2−4 2−4 2 = − 2 +10 2 − 5 = +1− 5 2 − 5 = 1− 5 2 − 5 −2 2 2 2 2 2ﺃﻱ ( )G = − 4 − 5 2 = − 2× 2 2 − 5 2 = −2 2 − 5 22ﻭ q∈Q a∈R ﺤﻴﺙ: a ﺕ : 02ﻜﺘﺎﺒﺔ ﻜل ﻤﻥ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﻌﻁﺎﺓ ﻋﻠﻰ ﺍﻟﺸﻜل: q( )( )3+2 3+( )( ) ( )3−5 3+2 = 3+ 5 = 9+3 5 +3 2+ 10 . 1ﻟﺩﻴﻨﺎ:5 3− 5 32 − 52 = 9+3 5+3 2+ 10 4 ﻭﺒﻨﻔﺱ ﺍﻟﻜﻴﻔﻴﺔ ﻨﺠﺩ ﺃﻥ : 2 = −1+ 3 .2 1+ 1 3 −1+ 5 = −17 15 − 34 5 .3 3−2 1 19 5 + 1 = 19 10 + 2 2 2 .4 ﺕ : 03ﻟﺩﻴﻨﺎ a = 11+ 3 7 :ﻭ b = 11− 3 7 .1ﺤﺴﺎﺏ a2 + b2ﻭ a × b ﻟﺩﻴﻨﺎ( ) ( )a2 + b2 = 11+ 3 7 + 11− 3 7 = 22 : ﻭﻟﺩﻴﻨﺎa × b = 11+ 3 7 × 11− 3 7 :
) = (11+ 3 7 )(11− 3 7 ( )= 112 − 3 7 2 = 121− 63 = 58 .22ﺍﺴﺘﻨﺘﺎﺝ (a + b)2ﻭ (a − b)2 ﻟﺩﻴﻨﺎ(a + b)2 = a2 + b2 + 2ab : = 22 + 2 58 ﻭﻟﺩﻴﻨﺎ(a − b)2 = a2 + b2 − 2ab : = 22 − 2 58 .3ﺍﺴﺘﻨﺘﺎﺝ ) (a + bﻭ )(a − b ﻟﺩﻴﻨﺎ(a + b)2 = 22 + 2 58 : ﻭﻋﻠﻴﻪa + b = 22 + 2 58 : ﻷﻥ a + b ≥ 0 : ﻭﻟﺩﻴﻨﺎ(a − b)2 = 22 − 2 58 : ﻭﻋﻠﻴﻪa − b = 22 − 2 58 : ﻷﻥa − b ≥ 0 : ﺕ : 04 .1ﺤﺴﺎﺏ ( )2 + 5 2 ﻟﺩﻴﻨﺎ ( ) ( )2 + 5 2 = 22 + 2× 2 5 + 5 2 : =9+4 5×9+4 5 .2ﺤﺴﺎﺏ 9 + 4 5 × 5 − 2 × 5 + 2 : ﻟﺩﻴﻨﺎ( ) ( )( )5 − 2 × 5 + 2 = 2 + 5 2 × 5 − 2 5 + 2 : =) (2+ 5 2 5−4 × =2+ 5
ﺕ : 05ﺍﻹﺜﺒﺎﺕ ﺃﻨﻪ :( ) ( )3− 7= 7+ 2 = 7− 1 = 1 3+ 7 6 7 +12 7−2 3 ( ) ( )3− 7 = 7 + 2 ﻟﺩﻴﻨﺎ : 2 7 −2 33+ 5ﻷﻥ 3(3 − 7 )(3 + 7 ) = 2( 7 − 2)( )7 + 2 :( ) ( )7 + 2 = 7 + 2 × 3 − 7 ﻭﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ 7 :33+ 7 33+ 7 3−= 7 −1 = 7 −1 × 7 +1 =1 6 6 7 +1 7 +1
ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ: -ﻤﻌﺭﻓﺔ ﺘﻌﺭﻴﻑ ﻭﺨﻭﺍﺹ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ -ﺍﻟﺘﺤﻜﻡ ﻓﻲ ﺍﻟﺤﺴﺎﺒﺎﺕﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ • ﺘﻌﺭﻴﻑ ,ﺘﺭﻤﻴﺯ • ﺨﻭﺍﺹ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ • ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ • ﺍﻟﺤﻠﻭل
• ﺘﻌﺭﻴﻑ .ﺘﺭﻤﻴﺯ :ﻭﺍﻟﻤﻌﺭﻑ ﻜﻤﺎ ﻴﻠﻲ a: ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ ﻟﻌﺩﺩ ﺤﻘﻴﻘﻲ aﻫﻲ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ﺍﻟﻤﻭﺠﺏ ﺍﻟﻤﺭﻤﺯ ﺇﻟﻴﻪ ﺒﺎﻟﺭﻤﺯ ﺇﺫﺍ ﻜﺎﻥ a = a : a ≥ 0 ﺇﺫﺍ ﻜﺎﻥ a = −a : a ≤ 0 ﻤﺜﺎل 129 = 129 :ﻷﻥ 129 ≥ 0 )− 35 = −(− 35 = 35ﻷﻥ 0 ≥ −35 • ﺨﻭﺍﺹ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ : • ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﺤﻘﻴﻘﻴﺎﻥ aﻭ a × b = a × b : b • ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ : a a2 = aﻭ − a = a −a = a 2 = a2a = a ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ,aﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻡ b •b b
ﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻡ ,aﻤﻬﻤﺎ ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺼﺤﻴﺢ ﺍﻟﻨﺴﺒﻲ : n • an = a n ﺃﻤﺜﻠﺔ : * 3×5 = 3 × 5 = 3x5 = 15 −3 = −3 * 7 7 = 3 7 * 12+(−5) = 7 =7 * 12 + − 5 = 12 + 5 = 17ﻤﻨﻪ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ ﻟﻤﺠﻤﻭﻉ )ﺃﻭ ﻓﺭﻕ( ﻋﺩﺩﻴﻥ ﻻ ﺘﺴﺎﻭﻱ ,ﺒﺎﻟﻀﺭﻭﺭﺓ ,ﻤﺠﻤﻭﻉ )ﺃﻭ ﻓﺭﻕ( ﻫﺫﻴﻥ ﺍﻟﻌﺩﺩﻴﻥ, ( )2 − 11 2 = 2 − 11ﻷﻥ 2 − 11 ≤ 0 )(= − 2 − 11 = −2 + 11
• ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ : .1ﺃﺤﺴﺏ ﺍﻷﻋﺩﺍﺩ C, B, Aﺍﻟﺘﺎﻟﻴﺔ :=B 3 − 3 + 6− 27 + 12 , =A 1−2 + 2− 2 − 1 − 5 2 3 4 2 C = −x + x2 +2x +1 )ﺤﻴﺙ xﻋﺩﺩ ﺤﻘﻴﻘﻲ( .2ﻟﻴﻜﻥ xﻋﺩﺩﺍ ﺤﻘﻴﻘﻴﺎ ﻭﻟﺘﻜﻥ ﺍﻷﻋﺩﺍﺩ D, C, B, Aﺤﻴﺙ, B = 1−2x + −1−2x +3 , A = 2x +1 + 2x −1 D = 2 − 3x − − 2 − 3x ، C = 3x+2 − 3x−2 +5 ﺒﺎﺴﺘﻌﻤﺎل ﺨﻭﺍﺹ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ ,ﺃﺤﺴﺏ ﻜﻼ ﻤﻥ A - Bﻭ C+D a .3ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﺤﻴﺙ a = 2 − 3 − 2 + 3 : (1ﻋﻴﻥ ﺇﺸﺎﺭﺓ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ a (2ﺃﺤﺴﺏ a2 (3ﺍﺴﺘﻨﺘﺞ ﻗﻴﻤﺔ ﻤﺒﺴﻁﺔ ﻟﻠﻌﺩﺩ a
• ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ:ﺕ : 01ﺤﺴﺎﺏ ﻜل ﻤﻥ ﺍﻷﻋﺩﺍﺩ C, B, A :ﺍﻟﻤﻌﻁﺎﺓ : =A −2 + 2 − 2 − 1 − 5 ﻟﺩﻴﻨﺎ : 3 4 2 = − (− 2)+ 2 − 2 + 1 − 5 3 4 2 = 2+ 2− 2 + 1 − 5 3 4 2 = 48 − 8 + 3 − 30 12 = 13 12 = 13 12B = 3 − 3 + 6− 27 + ﻭﻟﺩﻴﻨﺎ 12 : 2=) (− 3 2 − 3 + 6 − 27 + 12 = − 3 + 3 +6−3 3+2 3 2 = − 3 + 6 2 = 9 2 ﻭﻟﺩﻴﻨﺎC = − x + x2 + 2 x +1 : = − x + x 2 + 2 x +1
= − x + ( x +1)2 = − x + x +1 =1 : 02 ﺕ A − B ﺤﺴﺎﺏ B = 1− 2x + −1− 2x + 3 ﻭA = 2x +1 + 2x −1 :ﻟﺩﻴﻨﺎ A − B = 2x +1 + 2x −1 − 1− 2x − −1− 2x − 3 :ﻭﻋﻠﻴﻪ = 2x +1 + 2x −1 − − (2x −1) − − (2x +1) − 3 = 2x +1 + 2x −1 − 2x −1 − 2x +1 −3 = 0−3 = −3 C + D ﺤﺴﺎﺏC + D = 3x + 2 − 3x − 2 + 5 + 2 − 3x − − 2 − 3x= 3x + 2 − 3x − 2 + 5 + − (3x − 2) − − (3x + 2) : ﻟﺩ ﻴﻨﺎ= 3x + 2 − 3x − 2 + 5 + 3x − 2 − 3x + 2=5
ﺕ : 03ﻟﺩﻴﻨﺎ a = 2 − 3 − 2 + 3 : .1ﺘﻌﻴﻴﻥ ﺇﺸﺎﺭﺓ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ :a 2 − 3 − 2 + 3 2 − 3 + 2 + 3 a = 2− 3+ 2+ 3)= (2 − 3)− (2 + 3 2− 3+ 2+ 3= −2 3 2+ 3 + 2+ 3 ﻭﻟﻜﻥ − 2 3 ≤ 0 :ﻭ 2 + 3 + 2 + 3 ﻭﻤﻨﻪ aﺴﺎﻟﺏ ﺘﻤﺎﻤﺎ .2ﺤﺴﺎﺏ a2a2 = 2 − 3 − 2 + 3 2 ( ) ( )= 2 − 3 + 2 + 3 − 2 2 − 3 2 + 3)= 4 − 2 (2 − 3)(2 + 3 ﻟﺩﻴﻨﺎ:=4−2 1=2 .3ﺍﺴﺘﻨﺘﺎﺝ ﻗﻴﻤﺔ ﻤﺴﺒﻘﺔ ﻟﻠﻌﺩﺩ a ﻟﺩﻴﻨﺎ a2 = 2 :ﻭ aﺴﺎﻟﺏ ﻭﻋﻠﻴﻪa = − 2 :
ﺍﻟﻘﻴﻡ ﺍﻟﻤﻘﺭﺒﺔ ﻭ ﺍﻟﺘﻘﺩﻴﺭﺍﺕ ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ: -ﻜﺘﺎﺒﺔ ﻋﺩﺩ ﻋﺸﺭﻱ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻌﻠﻤﻲ -ﺘﻌﻴﻴﻥ ﻗﻴﻡ ﻋﺸﺭﻴﺔ ﻤﻘﺭﺒﺔ ﻟﻌﺩﺩ ﺤﻘﻴﻘﻲ -ﺘﻌﻴﻴﻥ ﻤﺩﻭﺭ ﻋﺩﺩ ﺤﻘﻴﻘﻲ -ﺘﻌﻴﻴﻥ ﺭﺘﺒﺔ ﻤﻘﺩﺍﺭ ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ • ﺍﻟﻜﺘﺎﺒﺔ ﺍﻟﻌﻠﻤﻴﺔ ﻟﻌﺩﺩ ﻋﺸﺭﻱ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ. • ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺸﺭﻴﺔ ﺍﻟﻤﻘﺭﺒﺔ ,ﺒﺎﻟﻨﻘﺼﺎﻥ ﺃﻭ ﺒﺎﻟﺯﻴﺎﺩﺓ ﻟﻌﺩﺩ ﺤﻘﻴﻘﻲ.• ﺘﻌﻴﻴﻥ ﺼﻴﻐﺔ ﻋﺩﺩ ﻋﺸﺭﻱ ﺒﺎﺴﺘﺨﺩﺍﻡ nﺭﻗﻤﺎ ﻋﺸﺭﻴﺎ ﻤﻥ ﺃﺭﻗﺎﻤﻪ. • ﻤﺩﻭﺭ ﻋﺩﺩ ﺤﻘﻴﻘﻲ. • ﺭﺘﺒﺔ ﻤﻘﺩﺍﺭ ﻋﺩﺩ ﻋﺸﺭﻱ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ. • ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ • ﺍﻟﺤﻠﻭل
• ﺍﻟﻜﺘﺎﺒﺔ ﺍﻟﻌﻠﻤﻴﺔ ﻟﻌﺩﺩ ﻋﺸﺭﻱ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ : ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ﻜل ﻋﺩﺩ ﻋﺸﺭﻱ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ Aﻋﻠﻰ ﺍﻟﺸﻜل : A = a x 10nﺤﻴﺙ aﻋﺩﺩ ﻋﺸﺭﻱ ﺒﺤﻴﺙ 1 ≤ a < 10ﻭ nﻋﺩﺩ ﺼﺤﻴﺢ ﻨﺴﺒﻲ ﻭﻋﻨﺩﺌﺫ ﺍﻟﻜﺘﺎﺒﺔ a x 10nﺘﺴﻤﻰ ﺍﻟﺸﻜل ﺍﻟﻌﻠﻤﻲ ﻟﻠﻌﺩﺩ .A ﻤﺜﺎل 0.0000712=7.12 x 10-5, 3250=3.25 x 103:• ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺸﺭﻴﺔ ﺍﻟﻤﻘﺭﺒﺔ ﺒﺎﻟﻨﻘﺼﺎﻥ ﺃﻭ ﺒﺎﻟﺯﻴﺎﺩﺓ ﻟﻌﺩﺩ ﺤﻘﻴﻘﻲ : ﺃ .ﻤﺜﺎﻻﻥ) :ﺤﻴﺙ ﺍﻟﺤﺴﺎﺒﺎﺕ ﺍﻷﻭﻟﻴﺎﻥ ﺘﻤﺎ ﺒﺎﻟﺤﺎﺴﺒﺔ( 2 3 = 0.666666 .... ≤ 0 .6 2 〈0 .7 ﺇﺫﻥ : 3ﻫﻲ 0.6 2 1 ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺸﺭﻴﺔ ﺍﻟﻤﻘﺭﺒﺔ ﺇﻟﻰ 3 10ﺒﺎﻟﻨﻘﺼﺎﻥ ﻟﻠﻌﺩﺩﻫﻲ 0.7 2 ﺒﺎﻟﺯﻴﺎﺩﺓ ﻟﻠﻌﺩﺩ 1 ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺸﺭﻴﺔ ﺍﻟﻤﻘﺭﺒﺔ ﺇﻟﻰ 3 10 7 =2.6457.5... ﺇﺫﻥ 2.64 ≤ 7〈2.65 :ﻫﻲ 2.64 7 ﺒﺎﻟﻨﻘﺼﺎﻥ ﻟﻠﻌﺩﺩ 1 ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺸﺭﻴﺔ ﺍﻟﻤﻘﺭﺒﺔ ﺇﻟﻰ 100ﻫﻲ 2.65 7 ﺒﺎﻟﺯﻴﺎﺩﺓ ﻟﻠﻌﺩﺩ 1 ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺸﺭﻴﺔ ﺍﻟﻤﻘﺭﺒﺔ ﺇﻟﻰ 100
ﺏ .ﺒﺼﻔﺔ ﻋﺎﻤﺔ: ﻟﻴﻜﻥ nﻋﺩﺩﺍ ﻁﺒﻴﻌﻴﺎ ﻭﻟﻴﻜﻥ Aﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻴﻭﺠﺩ ﻋﺩﺩ ﺼﺤﻴﺢ ﻨﺴﺒﻲ ﻭﺤﻴﺩ αﺤﻴﺙ: a ≤ 〈A a +1 10 n 10 na +1 1 aﻭﻋﻨﺩﺌﺫ 10 nﺘﺴﻤﻰ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺸﺭﻴﺔ ﺍﻟﻤﻘﺭﺒﺔ ﺇﻟﻰ 10nﺒﺎﻟﻨﻘﺼﺎﻥ ﻟﻠﻌﺩﺩ Aﻭ 10 nﺘﺴﻤﻰ 1 ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺸﺭﻴﺔ ﺍﻟﻤﻘﺭﺒﺔ ﺇﻟﻰ 10 nﺒﺎﻟﺯﻴﺎﺩﺓ ﻟﻠﻌﺩﺩ .A• ﺘﻌﻴﻴﻥ ﺼﻴﻐﺔ ﻋﺩﺩ ﻋﺸﺭﻱ ﺒﺎﺴﺘﺨﺩﺍﻡ nﺭﻗﻤﺎ ﻋﺸﺭﻴﺎ ﻤﻥ ﺃﺭﻗﺎﻤﻪ):(Troncatureﺼﻔﺔ ﺍﻟﻌﺩﺩ 35.672ﺒﺎﺴﺘﺨﺩﺍﻡ ﺭﻗﻤﻴﻥ ﻋﺸﺭﻴﻴﻥ ﻤﻥ ﺃﺭﻗﺎﻤﻪ ﻫﻲ ) 35.67ﺇﺤﺘﻔﺘﻀﻨﺎ ﺒﺭﻗﻤﻴﻥ ﺃﻤﺜﻠﺔ : ﺒﻌﺩ ﺍﻟﻔﺎﺼﻠﺔ(. •ﺼﻴﻐﺔ ﺍﻟﻌﺩﺩ 512.325ﺒﺎﺴﺘﺨﺩﺍﻡ ﺭﻗﻡ ﻋﺸﺭﻱ ﻭﺍﺤﺩ ﻤﻥ ﺃﺭﻗﺎﻤﻪ ﻫﻲ ) 512.3ﺍﺤﺘﻔﻀﻨﺎ ﺒﺭﻗﻡ ﻭﺍﺤﺩ ﺒﻌﺩ ﺍﻟﻔﺎﺼﻠﺔ(. •ﺼﻴﻐﺔ ﺍﻟﻌﺩﺩ 0.00357ﺒﺎﺴﺘﺨﺩﺍﻡ ﺜﻼﺜﺔ ﺃﺭﻗﺎﻡ ﻋﺸﺭﻴﺔ ﻤﻥ ﺃﺭﻗﺎﻤﻪ ﻫﻲ ) 0.003ﺍﺤﺘﻔﻀﻨﺎ • ﺒﺜﻼﺜﺔ ﺃﺭﻗﺎﻡ ﺒﻌﺩ ﺍﻟﻔﺎﺼﻠﺔ( •ﺼﻴﻐﺔ ﺍﻟﻌﺩﺩ 3.00001ﺒﺎﺴﺘﺨﺩﺍﻡ ﺃﺭﺒﻌﺔ ﺃﺭﻗﺎﻡ ﻋﺸﺭﻴﺔ ﻤﻥ ﺃﺭﻗﺎﻤﻪ ﻫﻲ ) 3ﺍﺤﺘﻔﻀﻨﺎ ﺒﺄﺭﺒﻌﺔ ﺃﺭﻗﺎﻡ ﺒﻌﺩ ﺍﻟﻔﺎﺼﻠﺔ ﻭ .(3.0000=3
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