Ma4 set-u-s54

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Ma4 set-u-s54

  1. 1. F F F F .4 F F F F FF 6 F 5 F (ก) () ก ก F {1, 4, 9, 16, 25, 36} Fก A F () () ก ก F {0, 1, 4} F ก (ก) - ( ) ก F2 FF 5 F 7 3 ก. 1 ∈ A . F x∈A F 1 ∈A x . F x∈A ก F 2x ∈ A ≡ x ∈ A ก F 2x ∉ A 1 ∈A ↔ 2 ∉ A 2 ∉A ↔ 4 ∈ A FF 1 4 ∈A 4 ∈A ↔ 8 ∉ A 8 ∉ A ↔ 16 ∈ A FF 1 16 ∈A 16 ∈ A ↔ 32 ∉ A 32 ∉ A ↔ 64 ∈ A FF 1 64 ∈AF 6 F 10 ก F S = {x / x 3 = 1} = {−1, 1} ก 1 : {x / x 3 = 1} = {1} ก 2 : {x / x 2 = 1} = {−1, 1} ก 3 : {x / x 3 = − 1} = {−1} ก 4 : {x / x 4 = x} = {0, 1} x4 − x = 0 x(x 3 − 1) = 0 x(x − 1)(x 2 + x + 1) = 0 x = 0, 1 1
  2. 2. F F FF 2 F 11 A = {x/x = 1 n n ก} = {1, 1 , 1 , 1 , .....} 2 3 4 A FF 6 F 12 4 x = ±3 x=2 A = {x / x 2 = 9 2x = 4} = {} , B = {x / x ∈ R x ≠ x} = {} C = {x / x ∈ R x 2 + 4 = 0} = {}, D = {x / x ∈ R x > x} = R / E = {x / x ก } = {}F 2 F 18 1 ก F C 3 2 F F B F F F A F Fก F B 3 F C F F Cก F A 1 2 4 F F C F ∴ ก 1 4F 2 F 20 2 1. F 2. ก F ก 3. F F F F 4.F 5 F 20 4 1. ก F ก 2. ก F F F 3. ก F F F 4. F F F F F 9 F 21 3 ก F A = {1, 2, {3}} B = {x / x = A} = {A} = {{1, 2, {3}}} C = {x / x ∈ A} = {1, 2, {3}} ก กF กF A=C 2
  3. 3. F F FF 12 F 22 a) F ก F F ∅ ก F F ก b) ก c) F ก ก F F d) ก e) ∅⊆∅ / f) A⊂B F B⊂A A=B F ก g) A⊂B F B⊂A F / ก A=B h) A⊂B a∈B F a F F ก A i) ก F A⊂B ก ก A F F BF 14 F 22 ก1 : F A ⊂ B, B ⊂ C F A⊂C F 3∉C FF 3∉A F ก2 :C C = {u, v, w, x, z} ก3 :C C = {u, v, w, x, z} ก4 กFF 16 F 23 A ก3 6! = 6 C 3 = 3!3! = 20F 17 F 23 กF F 1 F " ก " ก ก B⊂A B≠A B ก F3 = A − A ก0 − A ก1 − A ก2 − 1 = 2 6 − 6C0− 6 C1− 6 C2 − 1 = 64 − 6! − 6! − 6! − 1 = 41 6!0! 5!1! 4!2!F 3 F 24 B = {∅} n(B) = 1 21 2 n(P(P(P(B)))) = 2 2 = 22 = 2 4 = 16 3
  4. 4. F F FF 3 F 25 F F 4 {∅} A = {∅ , {∅} , { P(∅) }} P(A) = {∅ , {∅} , {{∅}} , {{{∅}}} , {∅ , {∅}} , {∅ , {{∅}}} , {{∅} , {{∅}}} , {∅ , {∅} , {{∅}}}F 5 F 25 1 ก P(A) Fก F F A = {0, 1 , 2} 2 ก 1 : 2x 3 − 5x 2 + 2x = 0 → x(2x 2 − 5x + 2 = 0) → x(2x − 1)(x − 2) = 0 → x = 0, 1 , 2 2 ∴ {x/2x 3 − 5x 2 + 2x = 0} = {0, 1 , 2} = A 2F 6 F 26 F F g) ก F A F P(A) FF FF A ก P(A) ก F F F h) F Fก B F n(B) ≤ n(A) F F F F A⊂BF 7 F 26 F ก2 F 2 3 F 8 F 27 A = {x / x ∈ N x < 100} = {1, 2, 3, 4, ....., 99} B = {x / x ∈ A 5 x } = {5, 10, 15, ....., 95} n(B) = 19 n(P(B)) = 2 n(B) = 2 19F 9 F 27 3 กF F F E = {a, b} F ก F E⊂A {a, b} ⊂ A ก3 F ก Fก AF 10 F 27 3 ก1 ก: F S = ∅, T = {∅} ก2 ก: F S = ∅ T = {1} U = {∅} ก3 ก4 ก 4
  5. 5. F F FF 11 F 28 a) P(B) ⊂ P(A) ก ก A B Fก F B⊂A FF P(B) ⊂ P(A) b) C F P(C) ≠ ∅ ก FF CF 12 F 28 n(P(A)) − n(P(C)) = 63 → 2 n(A) − 2 n(C) = 63 (1) n(P(A)) + n(P(B)) = 96 → 2 n(A) + 2 n(B) = 96 (2) ก (1) ก F ก F F ก 2 n(C) = 1 → n(C) = 0 F 2 n(C) = 1 (1) F 2 n(A) − 1 = 63 → 2 n(A) = 64 n(A) = 6 F 2 n(A) = 64 (2) F 64 + 2 n(B) = 96 2 n(B) = 32 → n(B) = 5 ∴ n(A) + n(B) + n(C) = 6 + 5 + 0 = 11F 2 F 29 ก1 ก {∅} ⊂ P(A) → ∅ ⊂ A ก ก2 ก {∅} ∈ P(A) → ∅ ∈ A ก ก3 {{a}, {a, b}} ⊂ P(A) → {a}, {a, b} ⊂ A กF F F F ก4 ก {{a, b}} ∈ P(A) → {a, b} ∈ A กF 3 F 29 ก A = {1, {1}, 2, {1, 2}, 3, {1, 2, 3}} 1. ก : {1, 2, 3} ∈ P(A) ก P F 1, 2, 3 ∈A ก 2. ก : ∅ ∈ P(A) P F ∅⊂A ก F F ∅ ∈ P(A) ก F ก ก F ∅ ⊂ P(A) ก P F ∅⊂A ก 5
  6. 6. F F F 3. : A ⊂ P(A) ก A = {1, {1}, 2, {1, 2}, 3, {1, 2, 3}} ⊂ P(A) ก P F 1 , {1} , 2 , {1, 2} , 3 , {1, 2, 3} ⊂ A F A F F ก F {A} ∈ P(A) ก P F A∈A F ก 4. ก : {{{1, 2}, 3}} ⊂ P(A) ก P F {{1, 2}, 3} ⊂ A ก F {1, 2}, 3 ∈ A กF 6 F 30 1. ก : ∅ ∈ P(P(B)) P F ∅ ⊂ P(B) ก 2. ก : ∅ ⊂ P(P(B)) ก F ก 3. : {∅} ⊂ P(B) / ก P F ∅⊂B / 4. ก : F F x=∅ F ∅ ∈ {∅} ∈ BF 7 F 30 (ก) ก {∅} ⊂ P(A) → ∅ ⊂ A ก () A ⊂ P(A) ก ก A=∅ F () ก {∅, A} ⊂ P(A) → ∅, A ⊂ A ก () {A} ∈ P(A) → ∅ ∈ ∅F 7 F 34 A B F ก 8 9 4 { }, {8}, {9}, {8, 9}F 8 F 35 4 ก F A = {1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10} ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 1 × 1 × 1 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 7 = 128 ก Fก X F 1, 2 F F 3 F X F 128 6
  7. 7. F F FF 9 F 35 กS = {X ∈ P(A) / 1 ∈ X n(X) ≥ 5} X⊂A F F X กA ก n(X) ≥ 5 F3ก n(X) = 5 n(X) = 6 n(X) = 7 ก n(X) = 5 F X F 6! 1⋅ 6 C 4 = 4!2! = 15 ก n(X) = 6 F X F 1⋅ 6 C 5 = 6! = 6 5!1! ก n(X) = 7 F X F 1⋅ 6 C 6 = 6! = 1 6!0! X Fก กก X ก ก ก ∴ X F = 15 + 6 + 1 = 22F 10 F 36 3 ก F {1, 2} ⊂ A F A F 1, 2 ก {1, 2, 4, 8, 16} ↓ ↓ ↓ ↓ ↓ 1 × 1 × 2 × 2 × 2 = 23 = 8 = 8 A F ∴ F = 8−1 = 7F 11 F 36 1 1 A ก3 ∴ n(P(A)) = 2 3 = 8 F P(A) ก F F 1 1+ 2+ 3+ 4+ C5 +8 C6 +8 C7 8C 8C 8C 8C 8 = 8! + 8! + 8! + 8! + 8! + 8! + 8! 7!1! 6!2! 5!3! 4!4! 3!5! 2!6! 1!7! = 8 + 28 + 56 + 70 + 56 + 28 + 8 = 254 2 A ก3 F P(A) ก F F 1 = P(A) − F ก − 1 = 2 8 − 1 − 1 = 254 7
  8. 8. F F FF 12 F 36 4 n(P(A)) = 32 → 2 n(A) = 2 5 → n(A) = 5 F B⊆A F ก B F F ก4 n(P(B)) ก = 2 4F 13 F 37 1 ก F P(B) = {∅, {0}, {{0, 1}}, { 0 {0, 1}}} ก x ∈ P(A) Fx⊂A F F x กA x ⊂ P(B) F ก x F F P(B) ก F A = {∅, {∅}, 0, 1} ↓ ↓ ↓ ↓ 2 × 1× 1 × 1 = 2 ∴ n(C) = 2F 14 F 37 a) A⊂X⊂B FF A⊂B F F F X F ก กF X = 0 b) A⊂X / X⊂B F ก X F ก กB F ∴ X = 2 5 = 32F 15 F 38 B⊂E F 1, 2, 3 F F E F E⊂A F 4, 5, 6, 7 F F F Eก F {1, 2, 3, 4, 5, 6, 7} ↓ ↓ ↓ ↓ ↓ ↓ ↓ 1 × 1 × 1 × 2 × 2 × 2 × 2 = 24 ∴ E F = 2 4 = 16 8
  9. 9. F F FF 16 F 38 X = {B ∈ P(A) / 1 ∈ B 2∈B 3 ∈ B} B⊂A F F B F 1 2 3 B ก A F F 1 A = {1, 2, 3, 4, 5, 6, 7} ↓ ↓ ↓ ↓ ↓ (2 3 − 1) × 2 × 2 × 2 × 2 = 7 × 2 4 = 112 ก F 1, 2, 3 F B ก 1, 2, 3 F B ∴ F X = 2 112 − 1F 17 F 38 X = {B ∈ P(A) / 1 ∈ B 2 ∉ B} B⊂A B F 1 F F 2 ก ก 1, 2 F B F B F 1 F 2 (ก F 1) A = {1, 2, 3, 4, 5, 6, 7, 8, 9} ↓ ↓ ↓ ↓ ↓ ↓ ↓ (2 2 − 1) × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 3 × 2 7 B F 3 × 10 7 ∴ n(X) = 3 × 10 7F 3 F 39 U = {0 , 2} A = {x / x 2 = 4} F A = {2} x = −2, 2 F x = −2 F F F F กU 9
  10. 10. F F FF 3 F 45 3 ก1 A ∩ B = {1, {1}} ≠ A ก2 A ∪ B = {∅, 1, {1}, {1, {1}}} ≠ A ก3 ก A ∩ B = {1, {1}} ∈ B ก4 A ∪ B = {∅, 1, {1}, {1, {1}}} ∈ BF 5 F 46 1. C ∩ D = {4} ⊂ A / 2. ก A − D = {3} C − B = {3} A−D ⊂C−B 3. A ∩ C = {3} (A ∩ C) ∪ B = {3, 4} ⊂ D / 4. A − D = {3} {3} ∈ A − DF 8 F 46 A − B = {1} B − A = {{1}} ∴ (A − B) ∪ (B − A) = {1 , {1}}F 9 F 47 ก1 A ∩ C = {2} ≠ ∅ ก2 ก A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9} (A ∪ B) = {} ก3 B ∩ C = B − C = {1, 9} (B ∩ C ) = (B − C) = {0, 2, 3, 4, 5, 6, 7, 8} ≠ 0 ก4 A − C = {0, 4, 6, 8} (A − C) = {1, 2, 3, 5, 7, 9} ≠ ∅ 10
  11. 11. F F FF 10 F 47 2 ก 1) B ∩ C = {3, 5, 7} 2) C ∩ B → C = {0, 1, 4, 6, 8, 9} ∴ C ∩ B = {1, 9} 3) C − B = {2} 4) A − B → A = {1, 3, 5, 7, 9} ∴ A −B = ∅ ∴ ก 2 = {1, 9}F 15 F 49 15.1 ก F ก F F F A B 7, 8 11 1 9 4,5 6 2, 3 10, 12 C U 15.2 1) A∪C = {2, 3, 4, 5, 7, 8, 9, 11} (A ∪ C) = {1, 10, 12} B − (A ∪ C) = {1, 4, 5, 6, 11} − {1, 10, 12} = {4, 5, 6, 11} 2) กF F [C ∪ (C ∩ A ∩ B)] ∩ [A ∪ (A ∩ B ∩ C )] ∩ [B ∩ (B ∪ C ∪ A )] =  C∪(A ∩ B ∩ C )  ∩  A ∪(A ∩ B ∩ C )  ∩  B ∪(A ∩ B ∩ C )        = (C ∩ A ∩ B ) ∪ (A ∩ B ∩ C ) = {2 , 3} ∪ {11} = {2, 3, 11} 11
  12. 12. F F FF 16 F 50 ก) I 6 ∪ I 12 = I 12 (I 6 ∪ I 12 ) ∩ I 8 = I 12 ∩ I 8 = I 8 ) I 20 − I 15 = {16, 17, 18, 19, 20} (I 20 − I 15 ) ∩ I 40 = {16, 17, 18, 19, 20} ) I 5 = {6, 7, 8, ...} I 9 = {10, 11, 12, ...} I 5 ∩ I 9 = {10, 11, 12, ...} = I 9 ) I7 ∪ I4 = I7 (I 7 ∪ I 4 ) ∩ I 10 = I 7 ∩ I 10 = ∅  (I 7 ∪ I 4 ) ∩ I 10  = ∅ = U  F 17 F 50 3 A n = [ 1 , 2n) n A 1, A 2, A 3, A 4 F A 1 = [1, 2) F A4 A2 = [ 1 , 4) A3 2 A2 A3 = [ 1 , 6) A1 3 1 1 1 1 2 4 6 8 A 4 = [ 1 , 8) 4 3 2 4ก F A1 ∪ A4 = [ 1, 1) 4 2 (A1 A4 ) - (A2 3 A 3) 3 A 2 3 A3 A 2 ∩ A 3 = [4, 8) A 1 A4 3 (A 1 ∪ A 4 ) − (A 2 ∩ A 3 ) = [ 1 , 1 ) ∪ [4, 8) 1 1 4 2 4 8 4 2 F 19 F 51 3F 20 F 51 2 F 1 F 60 1.1 ก Fก F x F d, e, f, g x⊂A F F x F กA A = { a, b, c, d, e, f, g} ↓ ↓ ↓ ↓ ↓ ↓ ↓ 2 × 2 × 2 × 1 × 1 × 1 × 1 = 23 ∴ n(S) = 2 3 = 8 12
  13. 13. F F F 1.2 a b F 22 − 1 = 3 b = {a, b} a − = {a} b = {b} − − = { } × ∴ A = { a , b, c, d, e, f, g} ↓ ↓ ↓ ↓ ↓ ↓ (2 2 − 1) × 2 × 2 × 2 × 2 × 2 = 3(2) 5 1.3 ก F S F a b c d n(S) = (2 4 − 1) ⋅ 2 3 = 15 ⋅ 2 3F 2 F 60 ก F S∩B ≠ ∅ F S F 1 2 F F S A A A = {1, 2, 3, a, b, c} ↓ ↓ ↓ ↓ (2 2 − 1) × 2 × 2 × 2 × 2 = 3 × 2 4 = 48 ∴ A S∩B ≠ ∅ F ก 48F 3 F 60 ก F A ∩ {1, 2, 3, 5, 7} = {2, 5, 7} A F 2, 5, 7 A F F 1, 3 ก A⊂ U F F A ก U ก F U = {1, 2, 3, 4, 5, 6, 7, ....., 18} ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 1 × 1 × 1 × 2 × 1 × 2 × 1 × 2 11 = 2 13 ∴ A F ก 2 13 F ก กF 13
  14. 14. F F FF 2 F 61 A∪B = {2, 3} P(A ∪ B) = {∅, {2}, {3}, {2, 3}} P(A) = {∅} P(B) = {∅, {2}, {3}, {2, 3}} P(A) ∪ P(B) = {∅, {2}, {3}, {2, 3}} P(A ∪ B) = P(A) ∪ P(B)F 4 F 62 กF F ก n(P(A)) − n(P(A)) = 63 n(P(A)) − n(P(B)) = 63 ก n(P(A)) − n(P(B)) = 63 2 n(A) − 2 n(B) = 63 F F ก 2 n(B) ก F 2 n(B) = 1 → n(B) = 0 F 2 n(A) − 1 = 64 → 2 n(A) = 64 → n(A) = 6 ∴ n(A) + n(B) = 6 + 0 = 6F 7 F 63 ก F ก P(A) ก B ∅, {∅}, {0, {0, 1}} n(P(A) ∩ B) = 3 P(A) B ก n(P(A) − B) = n(P(A)) − n(P(A) ∩ B) = 2 n(A) − 3 P(A) - B = 2 4 − 3 = 13F 9 F 63 ก F ก P(A) ก A ∅, {∅}, {0} n(P(A) ∩ A) = 3 A P(A) n(P(A) − A) = n(P(A)) − n(P(A) ∩ A) = 2 n(A) − 3 = 2 6 − 3 = 61 A P(A) n(A − P(A)) = n(A) − n(P(A) ∩ A) = 6 − 3 = 3 A P(A) n[(P(A) − A) ∪ (A − P(A))] = 61 + 3 = 64 14
  15. 15. F F FF 24 F 75 (A∩B) ∪ (A∩B ) ∪ (A ∩ B) = [A ∩ (B ∪ B )] ∪ (A ∩ B) U = A ∪ (A ∩ B) = (A ∪ A ) ∩ (A ∪ B) = U ∩ (A ∪ B) = A ∪ BF 25 F 76 ก. F กF B ∩C B ∪C [B ∪ ((A ∪ B ) ∩ (B ∪ C))] ∩ [B ∪ (B ∩ A)] ∩ (B ∪ A) = [B ∪ B ∪ (A ∩ C)] ∩ [B ∪ (B ∩ A)] ∩ (B ∩ A ) = U ∩ [(B ∪ B) ∩ (B ∪ A)] ∩ (B ∩ A ) = U ∩ B ∩ (B − A) = B − A . [(A ∪ B ∪ C) ∩ (A ∩ B) ∩ (A − B) ] ∪ [((A ∩ B) − C) ∩ (C − (A ∩ B ))] A∪B = (C ∩ A ∩ ∅ ) ∪ [∅ ∩ (C ∩ (A ∩ B ) )] = (C ∩ A ∩ U ) ∪ [ U ∩ ∅] = A ∪ ∅ = AF 26 F 76 (A ∆ B) = [(A − B) ∪ (B − A)] = [(A ∩ B ) ∪ (B ∩ A )] = [(A ∩ B ) ∩ (B ∩ A )] = [(A ∪ B) ∩ (B ∪ A)] = (A ∪ B) ∩ (A ∪ B ) F . ก = (A ∩ (A ∪ B )) ∪ (B ∩ (A ∪ B )) = ((A ∩ A) ∪ (A ∩ B )) ∪ ((B ∩ A) ∪ (B ∩ B )) = (∅ ∪ (A ∩ B )) ∪ ((B ∩ A) ∪ ∅) = (A ∩ B ) ∪ (B ∩ A) = (B ∩ A) ∪ (A ∩ B ) = (A ∩ B) ∪ (A ∩ B ) F ก. ก 15
  16. 16. F F FF 5 F 87 1 กF Fก F F ก F FF 3 1 ก 2 2 = 2 + 3 + 1 + 10 + 2 + 4 + 5 = 27 10 4 5 UF 6 F 87 F Humanities Sciences x ก F Humanities Social = 20 8-x 2 6 8 − x + 2 + 6 + x + 0 + 0 + 6 − x = 20 0 x 0 x = 2 6-x ∴ Humanities Sciences 2 Sciences UF 9 F 89 กF ก F F F = 35% = 48% F =x ก 35 - x x=12 48 - x 35 + (48 − x) + 29 = 100 = 23 = 36 x = 12 29 U = 100 1. = 35 + 36 = 71% 2. = 12% 3. F F = 23% 4. F = 36% 5. F = 35 + 29 = 64% 16
  17. 17. F F FF 16 F 92 กF ก F F F F F F ก1 = 65 + 9 + 6 + 7 F A F B 6 = 87 9 6 3 2 2 F F FFF 65 F F ก1 87 = 100 = 0.87 7 F C U = 100F 17 F 93 กF Fก F F ก F F F F 40 5 x 10 F 5 + x + 10 + 10 = 40 → x = 15 0 y z F 90 30 10 F 5 + y + 30 + 10 = 90 → y = 45 F U F F 70 F 10 + z + 30 + 10 = 70 → z = 20 ∴ 2 F F = x + y + z = 15 + 45 + 20 = 80F 18 F 93 กF Fก F F F U = 100% F F a + b + d + x ≥ 75 (1) a b c x b + c + e + x ≥ 70 (2) d e d + e + f + x ≥ 65 (3) f กF U = 100 17
  18. 18. F F F (1) + (2) + (3) : (a + b + c + d + e + f + x) +b + d + e + 2x ≥ 210% = 100% (b + d + e + x) + x ≥ 110% F 100% x ≥ 10% ∴ F F 10% 3F 19 F 94 3 กF Fก F F F ก F FF ก F ก F Fก F F F F F 3 ก F F ก F 1. Fก F ก F 2. ก F Fก F 3. F Fก 4. Fก กF 20 F 95 1 กF Fก F F Fก y F ≠ 0 z = 5y MATH ENG F y = 2 F z = 10 2x + 6 + y + z = 28 x x 0 2x + 6 + 2 + 10 = 28 y 6 z x = 5 0 F y = 4 F z = 20 2x + 6 + y + z = 28 HIS U = 100 2x + 6 + 4 + 20 = 28 x = -1 F F ∴ ก ก F F 5 18
  19. 19. F F FF 21 F 95 F n(A ∩ B ∩ C) = x กF ก F F F n(A ∪ B ∪ C) = 80 A B(42) 20 + 2 + (37 − x) + 5 + 3 + x + (20 − x) = 80 20 2 42-2-3-x 3 = 37-x กF ก F x = 7 5 x 28-5-3-x ∴ n(A ∩ B ∩ C) = 7 = 20-x C(28)F 25 F 97 n(P(A) ∪ P(B)) = n(P(A)) + n(P(B)) − n(P(A) ∩ P(B)) = n(P(A)) + n(P(B)) − n(P(A ∩ B)) = 16 + 8 − 4 = 20F 26 F 98 2 กF Fก F F ก = 100 U n(B ∪ A ) = 45 + 30 + 10 = 85 F 1 n(A − B ) = n(A ∩ B) = 30 F 2 ก 15 45 30 n(A ∪ B ) = n((A ∩ B) ) = 15 + 30 + 10 = 55 10 F 3 n(B ∩ A) = n(A ∩ B ) = 15 F 4F 27 F 98 n(P(A ∪ B)) = 2 n(A∪B) = 2 n(A) + n(B) − n(A∩B) n(P(A)) ⋅ n(P(B)) = 2 n(A) ⋅ 2 n(B) 2 n(A ∩ B) = n(P(A ∩ B)) (1) ก n(P(A) ∪ P(B)) = n(P(A)) + n(P(B)) − n(P(A) ∩ P(B)) z = x + y − n(P(A ∩ B)) n(P(A ∩ B)) = x+y−z F n(P(A)) , n(P(B)) n(P(A ∩ B)) (1) F n(P(A ∪ B)) = x + y − z xy 19
  20. 20. F F F F 28 F 98 2 n(x ∪ y ∪ z)= n(x) + n(y) + n(z) − n(x ∩ y) − n(x ∩ z) − n(y ∩ z) + n(x ∩ y ∩ z) 70 = 30 + 44 + 28 − n(x ∩ y) − 18 − 11 + 3 n(x ∩ y) = 6 F 29 F 99 F A = F F F F B = F F F F C = F F F F กF Fก F F A( F ) B( F ) a y b 3 x z c C( F ) U = 40n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C) 40 = 21 + 16 + 19 − (y + 3) − (x + 3) − (z + 3) + 3 x + y + z = 10 ก (a + b + c) + (x + y + z) + 3 = 40 a + b + c + 10 + 3 = 40 a + b + c = 27 ∴ F ก ก F = a + b + c = 27 F 31 F 99 ก n[(A ∪ C) ∪ (B ∪ C)] = n(A ∪ C) + n(B ∪ C) − n[(A ∪ C) ∩ (B ∪ C)] n(A ∪ B ∪ C) = 32 + 29 − n[(A ∩ B) ∪ C] n(A ∪ B ∪ C) = 32 + 29 − 28 = 33 ************************* 20

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