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

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SOLUCIONARIO DE ARITMETICA DE BALDOR

-16.

30 + 25 − 24

20

7

6 + 2 − 9 8

=

31 31

20

=

20

28 + 48 − 27 49

24 24

31

20 × 24

49 = 31 × 6

5 × 49 = 186

245

2 − 1⁄

3

8

5 + ⁄ 6

3

(5 ÷ 1⁄ 8

) × (1⁄ 5

÷ 1 ⁄

10

)

1 3⁄ 3

− 1⁄

3 + 5 8 6 × 1 1 2⁄

3

3

(5 × 8) × ( 1⁄ 5

× 10) = + 5 8 18

40 × 2

5

3 × 1 8 + 5 18

=

80

-17.

5

24 + 5 18

=

80

15 + 20

72

80

35

72 × 1 80 = 7

72 × 16 = 7

1 152

3⁄

4

1⁄ + 5 2⁄

3

1 6

⁄ 12

6 + (8 − 1⁄ 4

) + 3

17

3

4 × 6 + ⁄ 3

1⁄

12

6 + (7 4⁄ 4

− 1⁄ 4

) + 3

=

9

2 + 17

3 × 12 9

6 + 7 3⁄ + 3 =

2 + 68

4

13 3⁄ + 3

4

4 1 2 + 68

55⁄

4

35

72

80

+ 3 = 72 1 145

2

55⁄ + 3 = 2

55 4

⁄ + 3

4

145

2 × 4 55 + 3 = 29 × 2

11 + 3 = 58

11 + 3

-18.

5 3 11 + 3 = 8 3 11

1

8

1⁄ + 2 − ⁄ 2

1 4

⁄ 4

3 ÷ ( 5 3 × 6 = 8 × 4 + 2 − 1 2 × 4

5 ) 3 ÷ 2

32 + 2 − 2

3

2

= 32

3

2

32 × 2 3 = 64

3 = 21 1 3

-19.

1 + 1 2

3 + 1 − 1 3

2

2 1⁄

× (23

1 1 2

⁄ 2

÷ 47

12 )

− 3

5⁄

1 6

⁄ 6

Efectuando el numerador:

1 + 1 2

3 + 1 − 1 3

2

2

=

2 + 1 3

2

+

3

3 − 1 3

=

2

3 2

2

3 + 3

2

3

2 × 1 3 + 2 3 × 1 2 = 1 2 + 1 3 = 3 + 2 = 5 6 6

Efectuando el denominador:

2 1⁄

1 5

2

5⁄ − ⁄ 3

1 6

⁄ = 2

5 6 ⁄ − 1 3 × 6 = 5 2 × 6 5 − 2

6

= 3 − 2 = 1

Efectuando el paréntesis:

Tendremos:

-20.

23 1⁄ 2

÷ 47

12 = 47

2 × 12

47 = 6

5⁄

6

1 × 6 = 5 6 × 6 = 5

2 − 2 5

4⁄ + 3 − 1⁄

3

4 5

⁄ 3

4 − 1⁄

4

1⁄

+ 5 − 1⁄

× ( 7

24

5 20 × 11

2 )

2

Efectuando el numerador:

2 − 2 5

4⁄ + 3 − 1⁄

3

4 5

⁄ 3

= 1 5 5 − 2 5

4⁄

5

+ 2 3 3 − 1 3

4⁄

3

1 3 5

4⁄ + 2 2 8

3

4 5

⁄ = ⁄ 8 5

4 3 ⁄ + ⁄ 3

4 5

⁄ = 8 5 × 5 4 + 8 3 × 3 4

3

= 2 + 2 = 4

Efectuando el denominador:

4 − 1⁄

4

1⁄

2

+ 5 − 1⁄

5

24

3 4⁄ 4

− 1⁄

4

+ 4 5⁄ 5

− 1⁄

5

1⁄

24

2

3 3⁄

4

1⁄ + 4 4⁄

15

5

24 = ⁄ 24 4

1 2

⁄ + ⁄ 5

24 2

15

4 × 2 + 24

5 × 1 24 = 15

2 + 1 5

75 + 2

= 77

10 10

Efectuando el paréntesis:

Tendremos:

-21.

4

77⁄

10

7

20 × 11

2 = 77

40

× 77

40 = 4 × 10

77 × 77

40

40

77 × 77

40 = 1

1

1 − 1⁄ + 1

5

1 − 1⁄

6

× ( 1 1

1 − 1⁄ − 1

3

1 − 1 7 + 2 49 − 62

343 )

⁄ 8

Efectuando el numerador:

1

1 − 1⁄ + 1

5

1 − 1⁄

6

1

5⁄ 5

− 1⁄ + 1

6

5

⁄ 6

− 1⁄ = 1

4

6

⁄ + 1

5 5

⁄ 6

5

4 + 6 5

25 + 24

= = 49

20 20

Efectuando el denominador:

1

1 − 1⁄ − 1

3

1 − 1⁄

8

1

3⁄ 3

− 1⁄ − 1

8

3

⁄ 8

− 1⁄ = 1

2

8

⁄ − 1

7 3

⁄ 8

3

2 − 8 7

21 − 16

= = 5 14 14

Efectuando el paréntesis:

1

7 + 2 49 − 62

343

Tendremos:

49 + 14 − 62

= = 1

343 343

49⁄

20

× 1

5⁄

343 = 49

20 × 14

5 × 1

343

14

7

10 × 5 × 7 = 1 50

LEONARDO F. APALA TITO 175

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