estudiante

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
3 1 3
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
7
21
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
+ −
+ −
24 8 8 8 8 8

13. 1 –

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

14. 2 +

15.

(

3
5
1+
2

5+ 3

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

= 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 copyrightlaws as they
currently exist. No portion 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. (4x − 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)
4

3

2

3

2

2
y
+
6 y − 2 9 y2 −1

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)

=

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.
0

c.

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...