Solucionario purcell

Páginas: 749 (187203 palabras) Publicado: 27 de marzo de 2011
CHAPTER

0

Preliminaries
1 ⎡ 2 1 ⎛ 1 1 ⎞⎤ 1 8. − ⎢ − ⎜ − ⎟ ⎥ = − 3 ⎣ 5 2 ⎝ 3 5 ⎠⎦ ⎡ 2 1 ⎛ 5 3 ⎞⎤ 3 ⎢ − ⎜ − ⎟⎥ ⎣ 5 2 ⎝ 15 15 ⎠ ⎦ 1 ⎡ 2 1 ⎛ 2 ⎞⎤ 1 ⎡2 1 ⎤ = − ⎢ − ⎜ ⎟⎥ = − ⎢ − ⎥ 3 ⎣ 5 2 ⎝ 15 ⎠ ⎦ 3 ⎣ 5 15 ⎦ 1⎛ 6 1 ⎞ 1⎛ 5 ⎞ 1 =− ⎜ − ⎟=− ⎜ ⎟=− 3 ⎝ 15 15 ⎠ 3 ⎝ 15 ⎠ 9
2 14 ⎛ 2 ⎞ 14 ⎛ 2 ⎞ 14 6 ⎜ ⎟ = ⎜ ⎟ = ⎛ ⎞ 9. ⎜ ⎟ 21 ⎜ 5 − 1 ⎟ 21 ⎜ 14 ⎟ 21 ⎝ 14 ⎠ 3⎠ ⎝ ⎝ 3 ⎠ 2 2

0.1 Concepts Review
1. rationalnumbers 2. dense 3. If not Q then not P. 4. theorems

Problem Set 0.1
1. 4 − 2(8 − 11) + 6 = 4 − 2(−3) + 6 = 4 + 6 + 6 = 16 2. 3 ⎡ 2 − 4 ( 7 − 12 ) ⎤ = 3[ 2 − 4(−5) ] ⎣ ⎦ = 3[ 2 + 20] = 3(22) = 66 3.
–4[5(–3 + 12 – 4) + 2(13 – 7)] = –4[5(5) + 2(6)] = –4[25 + 12] = –4(37) = –148

=

14 ⎛ 3 ⎞ 2⎛ 9 ⎞ 6 ⎜ ⎟ = ⎜ ⎟= 21 ⎝ 7 ⎠ 3 ⎝ 49 ⎠ 49

2

⎛2 ⎞ ⎛ 2 35 ⎞ ⎛ 33 ⎞ ⎜ − 5⎟ ⎜ − ⎟ ⎜ − ⎟ 7 ⎠ = ⎝ 77 ⎠ = ⎝ 7 ⎠ = − 33 = − 11 10. ⎝ 6 2 ⎛ 1⎞ ⎛7 1⎞ ⎛6⎞ ⎜1 − ⎟ ⎜ − ⎟ ⎜ ⎟ ⎝ 7⎠ ⎝7 7⎠ ⎝7⎠ 7 11 – 12 11 – 4 7 21 = 7 7 = 7 = 7 11. 11 + 12 11 + 4 15 15 7 21 7 7 7 1 3 7 4 6 7 5 − + − + 5 12. 2 4 8 = 8 8 8 = 8 = 1 3 7 4 6 7 3 3 + − + − 2 4 8 8 8 8 8

4.

5 [ −1(7 + 12 − 16) + 4] + 2 = 5 [ −1(3) + 4] + 2 = 5 ( −3 + 4 ) + 2 = 5 (1) + 2 = 5 + 2 = 7

5.

5 1 65 7 58 – = – = 7 13 91 91 91 3 3 1 3 3 1 +− = + − 4 − 7 21 6 −3 21 6 42 6 7 43 =− + − =− 42 42 42 42
1 ⎡1 ⎛ 1 1 ⎞ 1⎤ 1 ⎡1 ⎛ 3 – 4 ⎞ 1⎤ = ⎜ – ⎟+ ⎜ ⎟+ 3 ⎢ 2 ⎝ 4 3 ⎠ 6 ⎥ 3 ⎢ 2 ⎝ 12 ⎠ 6 ⎥ ⎣ ⎦ ⎣ ⎦ 1 ⎡1 ⎛ 1 ⎞ 1⎤ = ⎢ ⎜– ⎟+ ⎥ 3 ⎣ 2 ⎝ 12 ⎠ 6 ⎦ 1⎡ 1 4⎤ = ⎢– + ⎥ 3 ⎣ 24 24 ⎦ 1⎛ 3 ⎞ 1 = ⎜ ⎟= 3 ⎝ 24 ⎠ 24

6.

13. 1 –

1 1 2 3 2 1 =1– =1– = – = 1 3 3 3 3 3 1+ 2 2

14. 2 +

3 5 1+ 2

= 2+

7.

3 3 = 2+ 2 5 7 − 2 2 2 6 14 6 20 = 2+ = + = 77 7 7

15.

(

5+ 3

)(

5– 3 =

) ( 5) – ( 3)
2

2

=5–3= 2

Instructor’s Resource Manual

Section 0.1

1

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing fromthe publisher.

16.

(

5− 3

) = ( 5)
2

2

−2

( 5 )( 3 ) + ( 3 )

2

27.

= 5 − 2 15 + 3 = 8 − 2 15

17. (3x − 4)( x + 1) = 3 x 2 + 3 x − 4 x − 4

= 3x2 − x − 4
18. (2 x − 3)2 = (2 x − 3)(2 x − 3)
= 4 x2 − 6 x − 6 x + 9 = 4 x 2 − 12 x + 9

4 2 + x + 2x x x + 2 12 4( x + 2) 2x = + + x( x + 2) x( x + 2) x( x + 2) 12 + 4 x + 8 + 2 x 6 x + 20 = = x( x + 2) x( x + 2) 2(3x + 10) = x( x + 2)
2

12

+

28.

2 y + 6 y − 2 9 y2 −1
2 y + 2(3 y − 1) (3 y + 1)(3 y − 1) 2(3 y + 1) 2y = + 2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1) = = = 6y + 2 + 2y 8y + 2 = 2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1) 2(4 y + 1) 4y +1 = 2(3 y + 1)(3 y − 1) (3 y + 1)(3 y − 1) 0⋅0 = 0

19.

(3x – 9)(2 x + 1) = 6 x 2 + 3 x –18 x – 9 = 6 x –15 x – 9
2

20. (4 x − 11)(3x − 7) = 12 x 2− 28 x − 33 x + 77
= 12 x 2 − 61x + 77

21. (3t 2 − t + 1) 2 = (3t 2 − t + 1)(3t 2 − t + 1)
= 9t − 3t + 3t − 3t + t − t + 3t − t + 1 = 9t 4 − 6t 3 + 7t 2 − 2t + 1
4 3 2 3 2 2

29. a. 22. (2t + 3)3 = (2t + 3)(2t + 3)(2t + 3)
= (4t 2 + 12t + 9)(2t + 3) = 8t 3 + 12t 2 + 24t 2 + 36t + 18t + 27 = 8t 3 + 36t 2 + 54t + 27

b.

0 is undefined. 0

c. e. 30. If

0 =0 17

d.

3 isundefined. 0

05 = 0

f. 170 = 1

23.

x 2 – 4 ( x – 2)( x + 2) = = x+2, x ≠ 2 x–2 x–2
x 2 − x − 6 ( x − 3)( x + 2) = = x+2, x ≠3 x−3 ( x − 3) t 2 – 4t – 21 (t + 3)(t – 7) = = t – 7 , t ≠ −3 t +3 t +3

24.

0 = a , then 0 = 0 ⋅ a , but this is meaningless 0 because a could be any real number. No 0 single value satisfies = a . 0 .083 12 1.000 96 40 36 4

31.

25.

26.

2x − 2x
3 2

2x − 2x + x

=

2 x(1 − x)
2

x( x − 2 x + 1) −2 x( x − 1) = x( x − 1)( x − 1) 2 =− x −1

2

Section 0.1

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without...
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