martes, 21 de julio de 2009

Actividad 13: ejercitación 1º año

• Suprimir paréntesis, corchetes, llaves y resolver. Cancelar de ser posible.

1. (3 - 8) + (- 5 - 2) - (- 9 + 1) - (7 + 5) =
2. (a + 3) + (15 - a) – (4 - a) – a =
3. 18 + [6 – 1 + (3 - 1)] - 2 =
4. [7 – ( x – 2)] – [2 – (x – 1)] =
5. [(a + b) – (c – d)] – [(a + b) + (d – c)] =
6. 49 + b – {5 + a + b + [14 – (b + 3 + a) – 9]} + 2 – c + [29 – (8 – c) =

• Calcular el valor de x que satisface las siguientes ecuaciones, (cancelar de ser posible).

1. x + (8 – 3) = (8 – 3) – 1
2. 4 – [5 – (6 + 3) – 2] = x – [4 – 2 + (5 – 7)]
3. 4 x – [(x + 5 – 2) + 6 – 3 ]= - [-(2 x – 5 + 6)]
4. 6 – [x – (3 – b) + 6 ]= 5 + (7 – b) – 2
5. 10 – {5 + [3 – (2 – x)]} = 0
6. 2 x – { - [-3 + (2 – x) + 6]} = - [5 – (-2) + (7 – 1)]
7. 8 – (7 – 2 x + 1) = x – { - 3 + [ 6 – (- 8 – 4) – 5]}

• Aplicar propiedad distributiva y resolver

1. 11 (m + 2) =
2. (a + b) 4 =
3. (x + 5) 8 =
4. 5 (x + 3) + (a + 8) 3 =
5. (2 – x) a + x (7 + a) =
6. (a + 5) (b + 2) =
7. (2 a + 3 b) (7 + 5 m) =
8. (a + b + c) ( m + n) =
9. (10 m + 8) (9 + 3 x) =
10. (4 r – 2 s) (x + y + z) =
11. (5 a – 3 b) (2 x + 4 y)=
12. (5 a b + 8) (x – 3 m) =

• Resolver

1. (3 – 4 . 5) (- 2) + 4 – 7 =
2. (5 – 2 . 7 + 2 – 6) . [(9 + 3 . 4 ) – (5 – 8)] + 3 . 3 + 4 =
3. – [ - (5 – 2) . (- 6) + (- 9) . (- 2)] – 9 (- 3) – (- 7 + 4). (- 1 – 1) =
4. [(- 6 + 4 + 3) (- 8 + 4) – (- 2)] (- 3 + 5) – (- 9 + 6) (- 8 + 5) – 2 =
5. – { - [( - 6 + 9) (- 2) – 8 (- 2)]} – (-3) . 2 =
6. – { [(-4) (- 3 – 1) + (- 7 + 5) (- 3 + 7)] . (- 2) + (- 7) (- 3)} + 1 =


• Hallar x

1. (- 3) (- 6) + 6 (x + 2) – [- 5 – 3 . 8] = 2 x + 3 (x + 5)
2. [4 – (- 3) . 1] (- 6 + 4) + 2 (x – 3) = x – [ 3 – (- 2) . 5]
3. (- 2) (- 1) + 5 (x – 6) + (2 – x) 2 = 2 (x + 3) – [- 2 – (- 3) (-1)]

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