Primera Practica

  • April 2020
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8) Dado los conjuntos A = {x ∈ R / (-x-3) ∈ [-2; 6>; B = {x ∈ R / 2x/5 ∈ [-10, 2]} C = {x ∈ R / -x ∈ A ↔ x ∉ B} Hallar (C-B) ∪ (A ∩ C) DESARROLLO En el conjunto A: -2 ≤ -x-3 < 6 -6 < x+3 ≤ 2 -9 < x ≤ -1 <-9; -1]

En el conjunto B: -10 ≤ 2X/5 ≤ 2 -50 ≤ 2X ≤ 10 -25 ≤ X ≤ 5 [-25 ; 5]

En el conjunto C: x∈A→x∈ B x ∈ Ac ∨ x ∈ B <-∞; 0> ∪ [3; ∞> ∪ [½; 7/2] HALLAR x∈A↔x∈C x ∈ (A ∩ C) ∨ x ∈ (Ac ∩ Cc) φ φ 0

½

3

7/2

13) Dado los conjuntos A = {x ∈ R / 2/2x+3 ∈ <¼; 2]}; b = {x ∈ R / x2/x-2 ≥ x+6} C = {x ∈ R / 3x2-4/x-6 ≤ x+6} HALLAR EL CONJUNTO M = {x ∈ R/ x → (C-A) → x → B} DESARROLLO En el conjunto A: ¼ < 2/2x+3 ≤ 2 ½ ≤ 2x+3/2 < 4 1 ≤ 2x+3 < 8 -2 ≤ 2x < 5 -1 ≤ x < 5/2 [-1; 5/2 >

En el conjunto C: 3x2 - 4/x-6 ≤ x+6 ( 3x + 2)( 3x - 2) ≤ (x+6)(x-6) ( 3x + 2) ( 3x - 2) ≤ 0 (x+6)(x-6)

En el conjunto B: x2/x-2 ≥ x+6 x2 - x(x-2) - 6(x-2)/x-2 ≥ 0 x2 - x2 -2x -6x + 12 ≥ 0 12x - 8x ≥ 0 4(3 - 2x) ≥ 0 3 - 2x ≥ 0 3 ≥ 2x x ≤ 3/2

-6 <-6; 6>

6

14) Dado los conjuntos A = {x ∈ R / x2 – 4 ≥ 0 ∧ x2 – 16 ≤ 0} B = {x ∈ R / x2 – 6 ≥ -x ∨ x2 –x <-1}; C = {x ∈ R/ (x2+x+1)>0 → (x2<x)} HALLAR EL CONJUNTO D = {x ∈ R / x ∈ (A-B) ↔ x ∈ (C-A)} DESARROLLO: En el conjunto A: x2 - 4 ≥ 0 x2 ≥ 4

-2



x2 - 16 ≤ 0 x2 ≤ 16

2

-4

4

A = <-∞; -2] ∪ [2; ∞> ∪ <-4; 4] En el conjunto B: x2 – 6 ≥ -x x2 + x – 6 ≥ 0 (x-2)(x+3) ≥ 0

-3 2 <-∞; -3] ∪ [2, ∞> En el conjunto C: x2 + x + 1 > 0 2 x + (½)2 > -1 + ¼ (x + ½)2 > -3/4 i (imaginario)



x2 – x < -1 x2 – x + 1 < 0 x2 – x + (½)2 < -1 + (½)2 (x - ½)2 < -1 + ¼ (x - ½)2 < -3/4 i (imaginario)

x2 < x x<1

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