Solucionario completo de Aritmetica de Baldor (Por Leonardo F. Apala T.)

SOLUCIONARIO DE ARITMETICA DE BALDOR
-2. (3 + 2)(4 + 5)
-3. (20 – 14)(8 – 6)
-4. (8 + 5 + 3)(6 – 4)
5 × 9 = 45
6 × 2 = 12
16 × 2 = 32
-5. (20 – 5 + 2)(16 – 3 + 2 – 1)
-6. (8 + 6 + 4)2
17 × 14 = 238
18 × 2 = 36
-7. (20 – 15 + 30 – 10)5
(5 + 20)5 = 25 × 5 = 125
-8. (50 x 6 x 42 x 18)9
(300 × 756)9
226 800 × 9 = 2 041 200
-9. (5 – 2)3 + 6(4 – 1)
3 × 3 + 6 × 3 = 9 + 18 = 27
-10. 3(8 – 1) + 4(3 + 2) – 3(5 – 4)
3 × 7 + 4 × 5 − 3 × 1
21 + 20 − 3 = 38
-11. (7 – 5)4 + 3(4 – 2) + (8 – 2)5 – 2(11 –
10)
2 × 4 + 3 × 2 + 6 × 5 − 2 × 1
8 + 6 + 30 − 2 = 44 − 2 = 42
-12. (11 – 4)5 – 4(6 + 2) + 4(5 – 3) – 2(8 –
6)
7 × 5 − 4 × 8 + 4 × 2 − 2 × 2
35 − 32 + 8 − 4 = 3 + 4 = 7
-13. (3 + 2)(5 – 1) + (8 – 1)3 – 4(6 – 2)
5 × 4 + 7 × 3 − 4 × 4
20 + 21 − 16 = 41 − 16 = 25
-14. (5 – 1)(4 – 2) + (7 – 3) (4 – 1)
4 × 2 + 4 × 3 = 8 + 12 = 20
-15. (3 – 2)(4 – 1) + 6(8 – 4) + (7 – 2) (9 –
7)
1 × 3 + 6 × 4 + 5 × 2
3 + 24 + 10 = 37
-16. 3(9 – 2) + 2(5 – 1) (4 + 3) + 3(6 – 4) (8
– 7)
3 × 7 + 2 × 4 × 7 + 3 × 2 × 1
= 21 + 56 + 6 = 83
-17. (8 – 2) 3 – 2(5 + 4) + 3(6 – 1)
6 × 3 − 2 × 9 + 3 × 5
18 − 18 + 15 = 15
-18. 300 – 3(5 – 2) + (6 + 1) (9 – 3) + 4(8 +
1)
300 − 3 × 3 + 7 × 6 + 4 × 9
300 − 9 + 42 + 36 = 291 + 78 = 369
-19. 500 + 6(3 + 1) + (8 – 5)3 – 2(5 + 4)
500 + 6 × 4 + 3 × 3 − 2 × 9
500 + 24 + 9 − 18 = 533 − 18 = 515
-20. 6[3 + (5 – 1)2]
6[3 + 4 × 2] = 6[3 + 8] = 6 × 11 = 66
-21. 8[(5 – 3)(4 + 2)]
8[2 × 6] = 8 × 12 = 96
-22. 9[(10 – 4)2 + (30 – 20)2]
9[6 × 2 + 10 × 2] = 9[12 + 20]
= 9 × 32 = 288
-23. [(5 + 2)3 + (6 – 1)5] [(8 + 6)3 – (4 –
1)2]
[7 × 3 + 5 × 5][14 × 3 − 3 × 2]
[21 + 25][42 − 6] = 46 × 36 = 1 656
-24. {15 + (9 – 5)2} {(6 x 4)3 + (5 – 4) (4 –
3)}
{15 + 4 × 2}{24 × 3 + 1 × 1}
{15 + 8}{72 + 1} = 23 × 73 = 1 679
-25. 800 + {20 – 3 x 4 + 5[18 – (6 – 1)3 + (5
– 2)4]}
800 + {20 − 12 + 5[18 − 5 × 3 + 3
× 9]}
800 + 8 + 5{18 − 15 + 12}
808 + 5 × 15 = 808 + 75 = 883
EJERCICIO 46
Efectuar, aplicando la ley distributiva:
-1. (8 + 3)2
-2. (7 – 5)3
8 × 2 + 3 × 2 = 16 + 6 = 22
7 × 3 − 5 × 3 = 21 − 15 = 6
-3. (9 + 6 – 2)5
-4. (b + c) a
-5. (x – y)m
9 × 5 + 6 × 5 − 2 × 5
45 + 30 − 10 = 65
b × a + c × a = ab + ac
x × m − y × m = mx − my
-6. (a + m – x)n
a × n + m × n − x × n = an + mn − nx
-7. 9(15 + 8 + 4)
9 × 15 + 9 × 8 + 9 × 4
135 + 72 + 36 = 207 + 36 = 243
-8. 7(25 – 18)
7 × 25 − 7 × 18 = 175 − 126 = 49
-9. 3(2 - 1 + 5)
3 × 2 − 3 × 1 + 3 × 5
6 − 3 + 15 = 18
LEONARDO F. APALA TITO 49