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

SOLUCIONARIO DE ARITMETICA DE BALDOR
-10.
11 + 15 + 10
( ) × 5 36
45 4
3 51 =
45 × 5 4
1 050
333
333
Se tiene:
9
9
= 333
1 050 1 050 = 111
350
333
0.18
0.6 + 0.1515 …
0.1010 … − 1 15
0.01818 …
0.18 = 18
100 = 9 50 ; 0.6 = 6 10 = 3 5 ;
0.1515 … = 15
99 = 5 10
; 0.1010 … =
33 99 ;
Ahora:
0.01818 … = 18 − 0
990 = 18
990 = 1 55
9
50
3
+
5
9 + 45 − 2
30
1
55
-11.
Se tiene:
5
33
10
99
1
55
=
− 1 15
52
30
1
55
=
3
10 + 3 2 − 1 15
1
55
= 52 52
× 55 =
30 6 × 11
26 286
× 11 =
3 3 = 95 1 3
3.2 − 2.11 … + 3.066 …
2.2 − 1.166 … + 2.033 …
3.2 = 3 2 10 = 32
10 = 16
5 ;
2.11 … = 2 1 9 = 19
9 ;
2.2 = 2 2 10 = 22
10 = 11
5 ;
1.166 … = 1 16 − 1
90
= 1 15
90 = 105
90 = 7 6 ;
3.066 … = 3 6 − 0
90 = 3 6 90 = 276
90 = 46
15 ;
2.033 … = 2 3 − 0
90 = 2 3 90 = 183
90 = 61
30
Ahora:
16
5 − 19
9 + 46
15
11
5 − 7 6 + 61 =
30
187
45
92
30
144 − 95 + 138
45
66 − 35 + 61
30
= 187
45 × 30
92 = 187
138 = 1 49
138
CAPITULO XXX POTENCIACION
EJERCICIO 190
Desarrollar, aplicando la regla anterior:
-1. (3 × 5) 2
-2. (2 × 3 × 4) 2
3 2 × 5 2 = 9 × 25 = 225
2 2 × 3 2 × 4 2 = 4 × 9 × 16 = 576
-3. (3 × 5 × 6) 3
3 3 × 5 3 × 6 3
27 × 125 × 216 = 729 000
-4. (0.1 × 0.3) 2
0.1 2 × 0.3 2 = 0.01 × 0.09 = 0.0009
-5. (0.1 × 7 × 0.03) 2
0.1 2 × 7 2 × 0.03 2
0.01 × 49 × 0.0009 = 0.000441
-6. (3 × 4 × 0.1 × 0.2) 3
3 3 × 4 3 × 0.1 3 × 0.2 3
27 × 64 × 0.001 × 0.008 = 0.013824
-7. (6 × 1 2 × 2 3 )2
(6 × 0.5 × 0.66 … ) 2
6 2 × 0.5 2 × 0.66 … 2
36 × 0.25 × 0.44 … = 4
-8. (2 × 0.5 × 1 5 )3
(2 × 0.5 × 0.2) 3 = 2 3 × 0.5 3 × 0.2 3
8 × 0.125 × 0.008 = 0.008
-9. (0.1 × 0.2 × 0.4) 4
0.1 4 × 0.2 4 × 0.4 4
0.0001 × 0.0016 × 0.0256
= 0.000000004096
-10. ( 1 4 × 4 × 1 2 × 6)4
(0.25 × 4 × 0.5 × 6) 4
0.25 4 × 4 4 × 0.5 4 × 6 4
0.00390625 × 256 × 0.0625 × 1 296
= 81
-11. ( 2 3 × 1 1 2 × 10 3 × 0.01)5
( 2 3 × 3 2 × 10
3 × 1
100 ) 5
( 2 3 ) 5
× ( 3 2 ) 5
× ( 10
3 ) 5
× ( 1
100 ) 5
2 5
3 5 × 35
2 5 × 105
3 5 × 15
100 5 = 10 5
3 5 × (10 2 ) 5
10 5
243 × (10 5 ) 2 = 1
243 × 10 5
1
243 × 100 000 = 1
24 300 000
-12. ( 5 6 × 1 1 5 × 0.3 × 6 2 3 )6
( 5 6 × 6 5 × 3 10 × 20
3 ) 6
( 5 6 ) 6
× ( 6 5 ) 6
× ( 3 10 ) 6
× ( 20
3 ) 6
5 6
6 6 × 66
5 6 × 36
10 6 × 206
3 6 = 206
10 6
EJERCICIO 191
Desarrollar:
-1. ( 1 2 )2 1 2
-2. ( 1 4 )2 1 2
64 000 000
1 000 000 = 64
2 2 = 1 4
4 2 = 1 16
-3. ( 5 7 )2 5 2
7 2 = 25
49
-4. ( 1 3 )3
LEONARDO F. APALA TITO 237