# Sol purcell

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• Publicado : 13 de septiembre de 2012

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CHAPTER

0

0.1 Concepts Review
1. rational numbers

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

2. dense

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

3. If not Q then not P.
4. theorems

2

ProblemSet 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

4.

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

5.

6.

7.

5 1 65 7 58
–=
–=
7 13 91 91 91
3
313
31+ −=
+−
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

Instructor’s Resource Manual

2

2
14 ⎛ 2 ⎞
14 ⎛ 2 ⎞
14 6

⎟= ⎜ ⎟= ⎛ ⎞
9.
⎜⎟
21 ⎜ 5 − 1 ⎟
21 ⎜ 14 ⎟21 ⎝ 14 ⎠
3⎠

⎝3⎠
2

=

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

⎛2
⎞ ⎛ 2 35 ⎞ ⎛ 33 ⎞
⎜ − 5⎟ ⎜ − ⎟ ⎜ − ⎟
7
⎠ = ⎝ 7 7 ⎠ = ⎝ 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
137 4675
−+
−+
5
12. 2 4 8 = 8 8 8 = 8 =
137 467 33
+−
+−
248 8888

13. 1 –1
1
2321
=1– =1– = – =
1
3
3333
1+ 2
2

14. 2 +

15.

(

3
5
1+
2

5+ 3

3
3
= 2+
25
7

22
2
6 14 6 20
= 2+ = + =
7777

= 2+

)(

) ( 5) – ( 3)

5– 3 =

2

2

=5–3= 2

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. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the 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)

12

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(3 x + 10)
=
x( x + 2)
2

+

= 4 x2 − 6 x − 6 x + 9
= 4 x 2 − 12 x + 9

19.

28.

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

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)
=

2

= 6 x –15 x – 9

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

=

21. (3t 2 − t + 1) 2 = (3t 2 − t + 1)(3t 2 − t + 1)
3

2

3

2

2

= 9t − 3t + 3t − 3t + t − t + 3t − t + 1

6y + 2 + 2y
8y + 2
=
2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)

=

= 12 x 2 − 61x + 77

4

2
y
+
6 y − 2 9 y2 −1

2(4 y + 1)
4y +1
=
2(3 y + 1)(3 y − 1) (3 y + 1)(3 y − 1)

= 9t 4 − 6t 3 + 7t 2 − 2t + 1

0⋅0 = 0

b.

0
is undefined.
0c.

0
=0
17

d.

3
is undefined.
0

e.

05 = 0

f. 170 = 1

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

23.

x 2 – 4 ( x – 2)( x + 2)
=
= x+2, x ≠ 2
x–2
x–2

24.

x 2 − x − 6 ( x − 3)( x + 2)
=
= x+2, x ≠3
x−3
( x − 3)

25.

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

26.

2

2x − 2x
3

2

2

x − 2x + x

=

2 x(1 − x)
2

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

Section 0.1

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

30. If

31.

.083
12 1.000
96
40
36
4

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