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

SOLUCIONARIO DE ARITMETICA DE BALDOR
100 3 = 1 000 000
101 3 = 1 000 000 + 3 × 100 2 + 3 × 100 + 1
= 1 000 000 + 30 000 + 300 + 1
EJERCICIO 203
Desarrollar:
-1.
-2.
-3.
-4.
-5.
-6.
-7.
-8.
= 1 030 301
(2 2 ) 2 = 2 2×2 = 2 4 = 16
(2 2 ) 3 = 2 2×3 = 2 6 = 64
(2 3 ) 4 = 2 3×4 = 2 12 = 4 096
(3 3 ) 4 = 3 3×4 = 3 12 = 531 441
[( 1 2 ) 2]
(1 3 ) 5 = 1 3×5 = 1 15 = 1
(5 2 ) 3 = 5 2×3 = 5 6 = 15 625
3
= ( 1 2 ) 2×3
= ( 1 2 ) 6
= 16
(0.01 2 ) 3 = 0.01 2×3 = 0.01 6
-9.
-10.
-11.
-12.
-13.
[( 1 4 ) 2]
2 6 = 1 64
= 0.000000000001
4
= ( 1 4 ) 2×4
= ( 1 4 ) 8
= 18
= 1
65 536
4 8
[(3 2 ) 3 ] 2 = 3 2×3×2 = 3 12 = 531 441
(a 3 ) x = a 3x
(x a ) 2 = x 2a
-14.
-15.
-16.
-17.
-18.
[(2 × 3) 2 ] 2 = 6 2×2 = 6 4 = 1 296
[(abc) 3 ] 4 = (abc) 3×4 = (abc) 12
= a 12 b 12 c 12
[( m 5 4]
n ) = ( m 4×5
n ) = ( m 20
n ) = m20
n 20
[(0.2 2 ) 2 ] 4 = 0.2 2×2×4 = 0.2 2×2×4
= 0.2 16 = 0.0000000000065536
[(0.3 2 ) 3 ] 2 = 0.3 2×3×2 = 0.3 2×3×2
= 0.3 12 = 0.000000531441
[( 2 3 2]
3 ) = ( 2 2×3
3 ) = ( 2 6
3 ) = 26
3 6 = 64
729
-19.
-20.
[( 3 5 ) 2]
2 2
[[( 2 2]
3 ) ]
EJERCICIO 204
-1.
-2.
3
= ( 3 5 ) 2×3
= ( 3 5 ) 6
= 36
5 6 = 729
15 625
= ( 2 2×2×2
3 ) = ( 2 8
3 )
= 256
6 561
= 28
3 8
2 2
3
(
5
) = (3 5 ) 9
6
5 ( 6 2 = 25
= 9
5 ) 36 36 = 1 4
25
( 0.2 × 2 3
1
2 × 1 )
3
0.008 × 8 27
1
8 × 1 =
27
3
= (0.2 × 2 3 ) 3
( 1 2 × 1 3 = 0.23 × ( 2 3
3 )
3 ) ( 1 3
2 ) × ( 1 3
3 )
8
1 000 × 8 27
1
8 × 1 =
27
8
125 × 1 27
1
8 × 1 27
-3.
8
125 × 8 = 64
125
( 22 × 3 5 × 4 2 2
2 4 × 3 2 ) = (22 × 3 5 × 4 2 ) 2
(2 4 × 3 2 ) 2
(2 2 ) 2 × (3 5 ) 2 × (4 2 ) 2
(2 4 ) 2 × (3 2 ) 2 = 24 × 3 10 × 4 4
2 8 × 3 4
-4.
16 × 59 049 × 256
256 × 81
16 × 729 = 11 664
[( ab 5 6
c ) ] = ( ab 5×6
c ) = ( ab 30
c ) = a30 b 30
c 30
-5.
-6.
[( ax 4 3
bm ) ] = ( ax 4×3
bm ) = ( ax 12
bm )
(ax) 12
(bm) 12 = a12 x 12
b 12 m 12
(2 2 ) 3 × (3 3 ) 2
(3 2 ) 3 × (2 3 ) 4 = 22×3 × 3 3×2
3 2×3 × 2 3×4 = 26 × 3 6
3 6 × 2 12
-7.
64 × 729
729 × 4 096 = 1 64
[ (2 3 ) 4
× ( 3 2 ) 2
2 × ( 1 3 ) 2 ] 2
=
[( 2 4
3 ) × ( 3 2 2]
2 )
[2 × ( 1 2 2]
3 )
[( 2 2 4]
3 ) × [( 3 2 2
2 ) 4×2
]
2
= (2 3 )
-8.
2 2 × [( 1 3 ) 2]
[(2 3 ) 3 ] 2
( 2 3 ) 8
× ( 3 2 ) 4
4 × ( 1 3 ) 4
=
× ( 3 2 ) 2×2
4 × ( 1 3 ) 2×2
2 8
3 8 × 34 2 4
2 4
4 × 1 =
3 4
4
3 4 3 4
= 24
4 = 16
4 = 4
(4 3 ) 2 = 23×3×2 218
43×2 =
4 6 = 218
(2 2 ) 6 = 218
2 12
= 2 6 = 64
LEONARDO F. APALA TITO 247