Pagina 027

Click here to load reader

  • date post

    20-Aug-2015
  • Category

    Documents

  • view

    201
  • download

    0

Embed Size (px)

Transcript of Pagina 027

  1. 1. Soluciones a las actividades de cada epgrafeSoluciones a las actividades de cada epgrafe1 Unidad 1. Divisibilidad y nmeros enteros PGINA 27 1 Calcula mentalmente. a) 5 7 b)2 9 c) 3 4 d)6 10 e) 5 12 f)9 15 g)12 + 17 h)22 + 10 i) 21 + 15 j) 3 6 k)1 9 l) 12 13 a) 2 b)7 c) 1 d)4 e) 7 f) 6 g) 5 h)12 i) 6 j) 9 k)10 l) 25 2 Resuelve. a) 10 3 + 5 b)5 8 + 6 c) 2 9 + 1 d)7 15 + 2 e) 16 4 6 f)22 7 8 g)9 8 7 h)15 12 + 6 a) 12 b)3 c) 6 d)6 e) 6 f) 7 g) 6 h)9 3 Calcula. a) 3 + 10 1 b)8 + 2 3 c) 5 + 6 + 4 d)12 + 2 + 6 e) 18 + 3 + 6 f)20 + 12 + 5 g)7 3 4 h)2 13 5 a) 6 b)9 c) 5 d)4 e) 9 f) 3 g) 14 h)20 4 Copia y completa como en el ejemplo. 7 4 6 2 + 5 + 3 4 = 15 16 = 1 a) 3 9 + 4 8 2 + 13 = 20 19 = 1 b)15 4 + 12 3 11 2 = 12 35 = 23 5 Calcula. a) 3 7 + 2 5 b)2 6 + 9 3 + 4 c) 7 10 5 + 4 + 6 1 d)6 + 4 3 2 8 + 5 e) 12 + 5 17 11 + 20 13 f)16 22 + 24 31 + 12 15 a) 7 b)6 c) 1 d)10 e) 4 f) 16 6 Quita parntesis y calcula. a) (3) (+4) (8) b)(5) + (6) (3) c) (+8) (+6) + (7) (4) d)(3) (+2) + (9) + (+7) a) (3) (+4) (8) = 3 4 + 8 = 1 b)(5) + (6) (3) = 5 6 + 3 = 2 c) (+8) (+6) + (7) (4) = 8 6 7 + 4 = 1 d)(3) (+2) + (9) + (+7) = 3 2 9 + 7 = 1 Pg. 1
  2. 2. Soluciones a las actividades de cada epgrafeSoluciones a las actividades de cada epgrafe1 Unidad 1. Divisibilidad y nmeros enteros 7 Resuelve de dos formas, como en el ejemplo. a) 10 (13 7) = 10 (+6) = 10 6 = 4 b) 10 (13 7) = 10 13 + 7 = 17 13 = 4 a) 15 (12 8) b)9 (20 6) c) 8 (15 12) d)6 (13 2) e) 15 (6 9 + 5) f)21 (3 10 + 11 + 6) a) 15 (12 8) = 15 (+4) = 15 4 = 11 15 (12 8) = 15 12 + 8 = 23 12 = 11 b)9 (20 6) = 9 (+14) = 9 14 = 5 9 (20 6) = 9 20 + 6 = 15 20 = 5 c) 8 (15 12) = 8 (+3) = 8 3 = 5 8 (15 12) = 8 15 + 12 = 20 15 = 5 d)6 (13 2) = 6 (+11) = 6 11 = 5 6 (13 2) = 6 13 + 2 = 8 13 = 5 e) 15 (6 9 + 5) = 15 (11 9) = 15 (+2) = 15 2 = 13 15 (6 9 + 5) = 15 6 + 9 5 = 24 11 = 13 f) 21 (3 10 + 11 + 6) = 21 (20 10) = 21 (+10) = 21 10 = 11 21 (3 10 + 11 + 6) = 21 3 + 10 11 6 = 31 20 = 11 8 Resuelve de una de las formas que ofrece el ejemplo: a) (8 13) (5 4 7) = (8 13) (5 11) = (5) (6) = 5 + 6 = 1 b) (8 13) (5 4 7) = 8 13 5 + 4 + 7 = 19 18 = 1 a) (4 9) (5 8) b)(1 6) + (4 7) c) 4 (8 + 2) (3 13) d)12 + (8 15) (5 + 8) e) (8 6) (3 7 2) + (1 8 + 2) f)(5 16) (7 3 6) (9 13 5) En cada caso, de cualquiera de las dos formas: a) (4 9) (5 8) = (5) (3) = 5 + 3 = 2 (4 9) (5 8) = 4 9 5 + 8 = 12 14 = 2 b)(1 6) + (4 7) = (5) + (3) = 5 3 = 2 (1 6) + (4 7) = 1 + 6 + 4 7 = 10 8 = 2 c) 4 (8 + 2) (3 13) = 4 (+10) (10) = 4 10 + 10 = 14 10 = 4 4 (8 + 2) (3 13) = 4 8 2 3 + 13 = 17 13 = 4 d)12 + (8 15) (5 + 8) = 12 + (7) (13) = 12 7 13 = 12 20 = 8 12 + (8 15) (5 + 8) = 12 + 8 15 5 8 = 20 28 = 8 e) (+2) (3 9) + (3 8) = (+2) (6) + (5) = +2 + 6 5 = 8 5 = 3 (8 6) (3 7 2) + (1 8 + 2) = 8 6 3 + 7 + 2 + 1 8 + 2 = 20 17 = 3 f) (11) (7 9) (9 18) = (11) (2) (9) = 11 + 2 + 9 = 11 11 = 0 (5 16) (7 3 6) (9 13 5) = 5 16 7 + 3 + 6 9 + 13 + 5 = 32 32 = 0 Pg. 2
  3. 3. Soluciones a las actividades de cada epgrafeSoluciones a las actividades de cada epgrafe1 Unidad 1. Divisibilidad y nmeros enteros 9 Resuelto en el libro de texto. 10 Calcula. a) 7 [1 + (9 13)] b)9 + [8 (13 4)] c) 12 [6 (15 8)] d)17 + [9 (3 10)] e) 2 + [6 (4 2 + 9)] f)15 [9 (5 11 + 7)] a) 7 [1 + (9 13)] = 7 [1 + 9 13] = 7 1 9 + 13 = 20 10 = 10 b)9 + [8 (13 4)] = 9 + [8 (9)] = 9 + [8 9] = 9 + [1] = 9 1 = 10 c) 12 [6 (15 8)] = 12 [6 15 + 8] = 12 6 + 15 8 = 27 14 = 13 d)17 + [9 (3 10)] = 17 + [9 (7)] = 17 + [9 + 7] = 17 + 16 = 1 e) 2 + [6 (4 2 + 9)] = 2 + [6 4 + 2 9] = 2 + 6 4 + 2 9 = 10 13 = 3 f) 15 [9 (5 11 + 7)] = 15 [9 (12 11)] = 15 [9 (+1)] = 15 [9 1] = = 15 [+8] = 15 8 = 7 11 Resuelve. a) (2 9) [5 + (8 12) 7] b)13 [15 (6 8) + (5 9)] c) 8 [(6 11) + (2 5) (7 10)] d)(13 21) [12 + (6 9 + 2) 15] e) [4 + (6 9 13)] [5 (8 + 2 18)] f)[10 (21 14)] [5 + (17 11 + 6)] a) (2 9) [5 + (8 12) 7] = (2 9) [5 + (4) 7] = (2 9) [5 4 7] = = (7) [5 11] = 7 [6] = 7 + 6 = 1 b)13 [15 (6 8) + (5 9)] = 13 [15 6 + 8 + 5 9] = = 13 15 + 6 8 5 + 9 = 28 28 = 0 c) 8 [(6 11) + (2 5) (7 10)] = 8 [(5) + (3) (3)] = 8 [5 3 + 3] = = 8 [8 + 3] = 8 [5] = 8 + 5 = 13 d)(13 21) [12 + (6 9 + 2) 15] = (13 21) [12 + 6 9 + 2 15] = = 13 21 12 6 + 9 2 + 15 = 37 41 = 4 e) [4 + (6 9 13)] [5 (8 + 2 18)] = [4 + (6 22)] [5 (10 18)] = = [4 16] [5 + 8] = 12 13 = 25 f) [10 (21 14)] [5 + (17 11 + 6)] = [10 21 + 14] [5 + 17 11 + 6] = = 10 21 + 14 5 17 + 11 6 = 35 49 = 14 Pg. 3