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

SOLUCIONARIO DE ARITMETICA DE BALDOR
-17. (2 ∙ 3 ∙ b) 5
2 5 ∙ 3 5 ∙ b 5 = 32 ∙ 243 ∙ b 5 = 7 776 b 5
-18. ( 15 3 )2 15 2
3 2 = 225
9 = 25
-19. ( a 6 )3 a 3
-20. ( 1 x )8 1 8
-21. ( 2×8
4 )2
6 3 = a3
216
x 8 = 1 x 8
2 2 × 8 2
4 2 = 4 × 64 = 4 × 4 = 16
16
-22. ( 3×6
9×2 )2
(3 × 6) 2
(9 × 2) 2 = 32 × 6 2
9 2 × 2 2 = 9 × 36
81 × 4 = 1
-23. ( ab
5cd )4
(ab) 4
(5cd) 4 = a4 b 4
5 4 c 4 d 4 = a4 b 4
625c 4 d 4
-24. ( 8×5×6
10×2×3 )2
(8 × 5 × 6) 2
(10 × 2 × 3) 2 = 82 × 5 2 × 6 2
10 2 × 2 2 × 3 2
64 × 25 × 36
100 × 4 × 9 = 64
4 = 16
Hallar por simple inspección, el resultado
de:
-25. 2 3 ∙ 5 3
-26. 50 4 ∙ 2 4
8 × 125 = 1 000
6 250 000 × 16 = 100 000 000
-27. 2 3 ∙ 5 3 ∙ 10 3
8 × 125 × 1 000 = 1 000 000
EJERCICIO 193
Desarrollar aplicando la regla anterior:
-1. (1 + 2) 2
1 2 + 2 × 1 × 2 + 2 2 = 1 + 4 + 4 = 9
-2. (6 + 9) 2
-3. (5 + 11) 2
-4. (12 + 15) 2
-5. (30 + 42) 2
6 2 + 2 × 6 × 9 + 9 2
36 + 108 + 81 = 225
5 2 + 2 × 5 × 11 + 11 2
25 + 110 + 121 = 256
12 2 + 2 × 12 × 15 + 15 2
144 + 360 + 225 = 729
30 2 + 2 × 30 × 42 + 42 2
900 + 2 520 + 1 764 = 5 184
-6. ( 1 2 + 1 3 )2
( 1 2 ) 2
+ 2 × 1 2 × 1 3 + (1 3 ) 2
= 1 4 + 1 3 + 1 9
-7. (0.5 + 3.8) 2
-8. (5 + 1 5 )2
9 + 12 + 4
36
= 25
36
0.5 2 + 2 × 0.5 × 3.8 + 3.8 2
0.25 + 3.8 + 14.44 = 18.49
5 2 + 2 × 5 × 1 5 + (1 5 ) 2
25 + 2 + 1 25 = 27 + 1 25 = 27 1 25
-9. (6 + 1 6 )2
-10. (0.1 + 5 6 )2
( 1 10 + 5 6 ) 2
1
100 + 1 6 + 25
36
6 2 + 2 × 6 × 1 6 + (1 6 ) 2
36 + 2 + 1 36 = 38 1 36
= ( 1 10 ) 2
+ 2 × 1 10 × 5 6 + (5 6 ) 2
9 + 150 + 625
= = 784
900 900
= 196
225
-11. (0.3 + 2 3 )2
( 3 10 + 2 3 ) 2
-12. (3 1 2 + 5 1 4 )2
= ( 3 10 ) 2
+ 2 × 3 10 × 2 3 + (2 3 ) 2
= 9
100 + 2 5 + 4 9
81 + 360 + 400
=
900
= 841
900
( 7 2 + 21 2
4 ) = ( 7 2
2 ) + 2 × 7 2 × 21
2
4 + (21 4 )
-13. (1 1 3 + 2 3 5 )2
= 49
4 + 147
4 + 441
16
= 196
4 + 441
16
=
784 + 441
16
= 1 225
16 = 76 9 16
( 4 3 + 13 2
5 ) = ( 4 2
3 ) + 2 × 4 3 × 13
2
5 + (13 5 )
16
9 + 104
15 + 169
25
-14. (8 1 2 + 3 4 )2
=
400 + 1 560 + 1 521
225
3 481 106
= 15
225 225
( 17
2 + 3 4 ) 2
= ( 17
2 ) 2
+ 2 × 17
2 × 3 4 + (3 4 ) 2
289
4 + 51
4 + 9 16 = 340
4 + 9 16 = 1 360 + 9
16
-15. (0.001 + 3
100 )2
1 369
16 = 85 9 16
( 1
1 000 + 3
100 ) 2
( 1
1 000 ) 2
+ 2 × 1
1 000 × 3
100 + ( 3
100 ) 2
1
1 000 000 + 3
50 000 + 9
10 000
1 + 60 + 900
1 000 000 = 961
1 000 000
-16. ( 3 5 + 1
10 )2
LEONARDO F. APALA TITO 239