Resolucao 2014 9ºano Algebra l1

7/23/2019 Resolucao 2014 9ºano Algebra l1

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Resoluções das Atividades

LIVRO 1 | ÁLGEBRA

SumárioCapítulo 1 – Potenciação................................................................................................................................................................................................................. 1

Capítulo 2 – Radiciação ................................................................................................................................................................................................................... 5

Capítulo 3 – Radicais ....................................................................................................................................................................................................................... 7

Capítulo 4 – Operações com radicais I ........................................................................................................................................................................................... 9

Capítulo 5 – Operações com radicais II ........................................................................................................................................................................................ 12

9o Ano – Ensino Fundamental II | 1

01 a) 34 = 3 · 3 · 3 · 3 = 81b) 83 = 8 · 8 · 8 = 512c) 2.0100 =1d) (–3)2 = (–3) · (–3) = + 9e) (–4)4 = (–4) · (–4) · (–4) · (–4) = + 256

f) 02010 = 0 0 0 0

2 010

0⋅ ⋅ ⋅ ⋅ =…

. fatores

g) −

= −

⋅ −

= +

1

7

1

7

1

7

1

49

2

h) (– 0,666...)2 =

x

x

x

x

= ⋅

=

− =

=

0 666 10

10 6 666

0 666

9 6

,

,

,

Logo, x x= =

÷

÷

6

9

2

3

3

3

( , )− = −

= −

⋅ −

= +0 666 2

3

2

3

2

3

4

92

2

i) (–0,111...)3

x = 0,111... · 10

10 1 111

0 1119 1

x

xx

=

− =

=

,

,,

Logo, x =1

9

02 a) 106 = 1.000.000b) 18 = 1

c) 023 = 0 d) (–1)2 = + 1 e) (–1)13 = – 1 f) (–100)12 = +1 000 0

24

. …

zeros

g) 10243 = 1 000 0243

. …

zeros

h) – (–1)194 = – (+1) = –1 i) – (–1)201 = – (–1) = + 1

( , ...)− = −

=

= −

⋅ −

⋅ −

=

−

0 111 1

9

1

9

1

9

1

9

1

7

3

2

229

03 I. a) ( , ) ( )− = −

= −

= − = −−− −

0 2 2

10

1

55 1253

3 3

2

b) − = − = −−6

1

6

1

362

2

c) ( )( )

− =

−

=−6

1

6

1

362

2

d) ( , )0 222 2

9

9

2

729

83

3 3

−−

=

= =

x

x

x

x

x

= ⋅

=

− =

=

=

0 222 10

10 2 2 22

0 222

9 2

2

9

,

,

,

e) − −

= − −

= − +

= −

−2

3

3

2

81

16

81

16

4 4

II. a)1

44

3

3=

− d) 19

10

110

19

1

10

19

10

191

1

= =

=

−

b)1

6

1

66

1

1= =

− e)1

77

9

9=

−

c)1

1010

5

5=

− f) 3

4

1

4

3

4

3

7

7

7 =

=

−

04 I. a) (–2)–3 – (–2)–2 – (–2)–5

−

− −

− −

−

− +

− −

1

2

1

2

1

2

1

8

1

4

1

32

3 2 5

− − + = − − +

= −1

8

1

4

1

32

4 8 1

32

11

32

b) (3 · 4)–1 · (3–1 + 4–1)–1

1

12

1

3

1

4

1

12

4 3

12

1

12

7

121

12

12

7

1

1

1

1

⋅ +

⋅ +

⋅

⋅ =

−

−

−

77

Testando seus Conhecimentos – pág. 4

Potenciação Aula 1Capítulo 1



2 | 9o Ano – Ensino Fundamental II

LIVRO 1 | ÁLGEBRA

01 a) 28.000 = 2,8000 · 10.000 = 2,8 · 104

b) 360 = 3,60 · 100 = 3,6 · 102

c) 0 03 3

100

3

103 10

2

2, = = = ⋅−

d) 0 00128 1 28

1 000

1 28

10 128 10

3

3, ,

.

,,= = = ⋅

−

e) 23,04 = 2,304 · 10

f) 128,03 = 1,2803 · 100 = 1,2803 · 102

g) 346,15 = 3,4615 · 100 = 3,4615 · 102

h)0 006937

6 937

1 000

6 937

10 6 937 103

3

,

,

.

,

,= = = ⋅

−

02 a) 1,5 · 100.000.000 = 1,5 · 108

b) 3,84 · 100.000 = 3,84 · 105

01 I. a) ( ) ( ) ( )

( )( )

− ⋅ −

−( )

= −

− = −

5 5

5

5

55

3 38

2 20

41

40

1

b) ( ) ( ) ( ) ( )

( )

( )

( )

− ⋅ −

⋅

=

− ⋅ −[ ]⋅

=+

+

2 10

4 5

2 10

4 5

20

20

50 50

37 37

50

37

50

37 == +( )20 13

c) ( ) ( )− − − +

− −

− +

− −

− − − − −

−

2 2 2 2

1

2

1

4

1

2

1

4

2 1

4

1 2 2 1 2 2

2 2

−

+

−

−

+ − = + −

−

−

2 2

2 2

2 1

4

3

4

3

4

16

9

9

16

256 81

144 ==

+175

144

d)

3

2

2

3

2

3 3 2

2

3

2

3

5 4

10

1 3

5

÷

÷ ⋅

=

÷

− −

−( )

÷ ⋅

=

÷

∴

∴

4

10 3

9

10 32

3

1

3 2

2

3

2

3

2

3

2

33

2

3

2

3

4

9

9

7

2

=

=

II. a) 1,32 = (1,3) · (1,3) = 1,69 b) 0,42 = (0,4) · (0,4) = 0,16 c) –0,26 = –(0,2) · (0,2) · (0,2) · (0,2) · (0,2) · (0,2) =

– 0,000064 d) (–0,08)3 = (–0,08) · (–0,08) · (–0,08) = – 0,000512

05 2 2 2 1 2 3

16

1

2 010

1

2

1 1 2 2

1 1

− − −− −

+ − − − ÷ − +

+ −

( ) ( ) ( ).

−− −

÷ −

+ −

−[ ] ÷ −

+ −

1

2 4 1

1

4

16

3 2 010

4

4 1

4

16

3 2 010

.

.

−−[ ] ÷ +

+ −

−[ ] ÷ +

+ −

− + −

4 3

4

16

3 2 010

4 4

3

16

3 2 010

16

3

16

3

.

.

22 010 2 010. .⇒ −

Atividades Propostas – pág. 5

c) 9 11

100 009 11 10

31

31,,

…

zeros

= ⋅−

d) 5 974 1 00 0 5 974 1024

24, ,⋅ = ⋅…

zeros

e) 1 66

100 0

166 10

23

23,,

…

zeros

= ⋅−

f) 3 · 100.000.000 = 3 · 108

03 x y

xy

x y

xy

y x

xy

xy

x y

xy

xyy x

− −

−

+=

+

=

+

=+

⋅ = +

1 1

1

1 1

1 1 1( )

04 a) − − − −

− − − −

− − − − − − +[

2 3 2 3

8 9 8 9

8 9 1

8 9 1

3 2 3 2 51

51

51

( )

( )

( )

]]

− − = −8 10 18

b) 2

5

5

42 3

5

2

5

4

1

49

25

4

2

2 2

2

÷ − − − −

÷ − +

− +

+

−−( ) ( )

( )

⋅ +

− −

+ − −

− − = − −

= −

4

5

1

49

5

1

4 9

1

4

4

1

1 16

4

17

4

05 E x y x y

x y x y

x y x

x y y

x

y

E

= −

+ =

⋅ −⋅ +

= −

+

=−

3 3 4 3

3 3 3 4

3 3

3 3

1

1

1

1

1 1

2

( )

( )

+ − =

−−

= −−

= +

−1

1 3

1 2

1 3

1

2

1

2( )

06 C, F

Escrevendo o valor em função da potência de base dez,tem-se o seguinte resultado: 3,75 · 10–8, que, por sua vez,pode ser representado como 0,375 · 10–7 (item C) ou375 · 10–10 (item F), os quais determinam o mesmo resul-tado da expressão dada na questão.





LIVRO 1 | ÁLGEBRA

c) ( , )

( )

0 2

5

2

10

1

5

1

5

1

5

21

1 15

21

15

21

− − − −=

=

115

36

361

55=

= −

d) ( )x xn n3 2 6− −

=

II. a b

c

c

a b

a b

c

c

a

b b

c

2 3

3

2 2

6 3

3

2

6

2 5

1

1

3 1

⋅

⋅

⋅⋅ ⋅ = =

⋅

−

( 00

2 10

3 10

2 10

3

2 10

243 100

224 300

2 12 15

8 5

38

5 40

38

52

)

..

⋅ =

= ⋅

⋅ = ⋅ =

⋅=

= = 00

02 a) 2 10 7 3 2 1 0 7

10 7 2 10 7

8 10 7

3 10 7

8

3

⋅ −

+ ⋅ ⋅ −

− + ⋅ −

=⋅

−

⋅ −

=

b)( ) ( )

( ) ( )

10 3 3 10 3 2 3

10 2 3 10 4 4

10 9 1018

10 6 10 16

− ⋅ −

−

− ⋅ − =

− ⋅− ⋅ −

==−

=109

10 22 1031

03 ( ) ( )

( ) ( )

( ) (10 2 5 10

2 10 5

10 2 55 4 2 3 2 3

3 2 2

5 4 2− − − −

−

− −÷ ÷ ÷− ⋅ − ÷

= ⋅ ÷ 33 2 3

10 8

9 6

10

1

8 100 25

10 2

5 10

100

8

⋅

−

⋅ +

÷

∴

⋅⋅

−

−

−

)

( )

÷

=

⋅ ⋅⋅ ⋅

− ⋅⇒

⋅⋅

−=

− −

−

− −

−

25

1

2 5 2

5 2 5

100

8

1

25

2 5

2 5

1

2

10 10 8

9 6 6

2 10

6 3 22 5

2

2 5 2 5 10

8 7

1

7 7 7 7

− −

−

− − − −

⋅−

∴

− ⋅ = − ⋅ = −( )

04 a) (V)

b) (F) (+3 + 4)2 = (+7)2 = 49, logo, falso 32 + 42 = 9 + 16 = 25

c) (V)

d) (F) 4 3 1 024

27

4

3

16

9

1 024

27

16

9

5 3

2

÷ =

=

≠

.

.

e) (F) (a – b)2

= a2

– 2ab + b2

≠ a2

– b2

f) (F) a

b

a

ba b

= ≠ −

2 2

2

2 2

05 − ⋅ − ⋅ − −( )

− −( ) ⋅

∴

∴

−

− −

− −

0 25 0 125 0 0625

0 5 0 03125

4 2 3 2 3

3 5

, ( , ) [ , ]

, ,

[(00 5 0 5 0 5

0 5 0 5

2 4 3 2 3 4 2 3

3 5

, ) ] { [( , ) ] } { [ , ] }

( , ) [( , ) ]

⋅ − ⋅ − −( )− − ⋅

− −

− −55

8 18 24

3 25

34

0 5 0 5 0 5

0 5 0 5

0 5

∴

∴− ⋅ ⋅ −

+ ⋅

∴

∴−

− −

− −

−

( , ) ( , ) ( , )

( , ) ( , )

( , )(( , )

( , )0 5

0 528

6

−

−

= −

01 a) 33p · 3–2p = 33p + (–2p) = 3p

b) ( ) ( ) ( ) ( ) ( ) ( ) ( )− ⋅ − ⋅

= − ⋅ − ⋅ + = − ⋅ + = −−2 2

1

2 2 2 2 2 2 24

6

4 1 6 5 6 11

c) 52x + 1 ÷ 5x + 1 = 52x + 1 – (x + 1) = 52x – x + 1 – 1 = 5x

d) [ ( ) ] ( )

[ ( ) ] (

4 8 125 10 2 25

2 2 5 10 2

5 2 14 5 11 1 4

10 6 3 14 20 44

⋅ ÷ ⋅ ⋅ ∴

∴ ⋅ ⋅ ÷ ⋅ ⋅

−

55

2 2 5 10 2 5 2 5

2 5 2 5

8

10 84 4 2 20 44 894 42

20 20 44 8

−

−

−

∴

∴ ⋅ ⋅ ÷ ⋅ ⋅ ⇒⋅

⋅ ⋅ ⋅

)

[ ] ( ) ∴∴

∴⋅

⋅

= ⋅ = ⋅ =2 5

2 5 2 5 2 5 10

94 42

64 12

30 30 30 30( )

e)[( ) ]

[ ] [ ]( )a

a a

a

a

a

aa a

p p pp p

−

−

−

−

−

−

− + − ⋅ −

⋅

= = = =

3 2

4 10 5

6

6 5

6

30

6 30 6 5

02 a) −( ) +( ) = −

−( ) +( )

4 4 47 7 14

.

b) − − ⋅ + = −

− +

( ) ( )( ) ( )

3 3 310 13 23

c)− − ⋅ − ⋅ − = +

− + −

( ) ( ) ( )( ) ( ) ( )

2 2 2 218 20 21 59

d) − −

÷ − +

− +

1

5

1

5

3 5 5 4

( ) ( )

= −

÷

= −

−1

5

1

5

1

5

15 20 5

03 m m=

⋅ ⋅

⋅

= = ∴ = =

− −

− −

−

−

10 10 10

10 10

10

10

1 1 15 6 5

3 1 3

6

6

2 2

( )

( )

04 D

a2 = 996, b3 = 997 e c4 = 998

(abc)12 = a12 · b12 · c12

= (a2)6 · (b3)4 · (c4)3

= (996)6 · (997)4 · (998)3

= 9936 · 9928 · 9924 = 9988

01 P = 3 · (–1)2n + 5 · (–1)2n + 1 – (–1)2n + 6 + 4 · (–1)3n

a) n e ímpar: P = 3 · (+1) + 5 · (–1) – (+1) + 4 · (–1) + 3 – 5 – 1 – 4 = – 7 b) n é par: P = 3 · (+1) + 5 · (–1) – (+1) + 4 · (+1) + 3 – 5 – 1 + 4 = +1

02 x = (–1)8n – (–1)4n + 7 + (–1)7n – 2 · (–1)n Considerando n par:

x = (+1) – (–1) + (+1) – 2 · (+1) + 1 + 1 + 1 – 2 = + 1 Considerando n ímpar: X = (+1) – (–1) + (–1) – 2 · (–1) + 1 + 1 – 1 + 2 = + 3

03 a)2 2 2 2

2 2 2

2 2 2

2 2 1

8 2

1

2 1

43 1

1

3 1

1

n n

n n

n

n

⋅ − ⋅

⋅ +

=

−( )+( )

=−

+

=− −






LIVRO 1 | ÁLGEBRA

01 a)

( ) [( ) ( ) ]

( )

( ) (10 10 10 10 10

10 10

10 10 102 2

2

2x y x x y y

x y

x x y+ ⋅ − +

+=

− ++ + yy

x y

)2

10 10+

b) 2 1 2 5 2

2 1 2 17

3 20

1 8 17

17

24

17

24

2

3

x

x

⋅ + − ⋅

⋅ − −

=−

− −

=−

−

=+( )

( )

02 a) 10 10

10

10 10 1

10 10

1000 1

100

999

100 9

2 3 2

2 2

2 3

2 2

a a

a

a

a

+

+

−=

⋅ −

⋅

=−

= =( )

,999

b) a b

a ba b

−( )

−( )= −

3

2

03 E a ba b

a b=−

−

= − = ⋅ − ⋅ = ⋅ ⋅ − ⋅ =

⋅ − ⋅

( )( )

3

2

8 6 2 6 6

6

5 10 2 10 5 10 10 2 10

500 10 2 100 500 2 10 498 10 4 98 106 6 6 8= − ⋅ = ⋅ = ⋅( ) ,

04 2 2 2 2

2 2 2

20 1

21

4

1

2

40 1

21 2

4

41

23

4

41

2

4 2 1

2 1

n

n

⋅ + +

⋅ +

=

+

+

=

+

+= = ⋅

−

− −

( )

( )

44

3

82

3

2

=

05 E

1

2

1

2

1

22 2

1

2

1

2

1

2

4

4

4

− ⋅

⋅

=

⋅

−

+

− −

x x

x

x

⋅

1

2

1

2

4x

07 Basta contar os fatores 5.

⇒ 5, 10, 15, ... , 1.000⇒ Tem 200 fatores. ⇒ 25, 50, 75, ... , 1.000 ⇒ Tem 40 fatores.

⇒ 125, 250, ... , 1.000⇒ Tem 8 fatores.

⇒ 625⇒ 1

249

fator

fatores

Logo⇒ 249 zeros.

08 ⇒ 5, 10, 15, ... , 100 ⇒ Tem 20 fatores. ⇒ 25, ... , 100 ⇒ Tem 4 fatores.

Logo, 24 zeros, portanto a maior potência de 10 quedivide P é 1024.

09 1 + 2 + 1 = 4 = 22

1 + 2 + 3 + 2 + 1 = 9 = 32

1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 = 42

A = 1 + 2 + 3 + ... + 2.008 + 2.009 + 2.008 + ... + 1 = 2.0092

A

287

2 009

287

2 009

287 7 49

2

2

2

2

2= =

= =

( . )

( )

.

10 Basta contar os fatores 5. 5, 10, 15, ... , 200⇒ Tem 40 fatores. 25, 50, ... , 200⇒ Tem 8 fatores. 125⇒ 1 fator.

Logo, 49 fatores, portanto, 49 zeros.

06 C 2x = b

2 2 2

1

42

1

4

1

4 42 2 2 2 2 2 2

2− + −

= ⋅ = ⋅ = ⋅ = ⋅ =x x x b b

b( ) ( )

b) 5 5 5 5 5 5 5

5 5 5

5 5 5 5

5 1 5

95

6

2 2 2 2 2 3

2 2

2 2 3

2

n n n n

n n

n

n

⋅ + ⋅ − − ⋅

+ ⋅

=

+ −( )+( )

= −

c) 2 2 2 2 2 2

13 2 2

2 2 2 2

2 13 2

7

13

3 3 3 3 2

3 1

3 3 2

3

n n n

n

n

n

⋅ + ⋅ + ⋅

⋅( ) =

+ +( )⋅

=

04 a) 3 1

81

3 1

33

4

4

4

x

x

x

=

= =

= −

−

d) 10x = 0,00000001 10x = 10–8

x = – 8

b) 10x = 1.000.000 e) (0,001)x =1

1093

10x =106 (10–3)x = 10–93

x = 6 10–3x = 10–93

– 3x = –93 c) 5x = 0,2 =

2

10

1

5= x = 31

5x = 5–1⇒ x = –1

05 E x y x yx y x y

x y x

x y y

E x

y

= −+ = ⋅ −⋅ +

= −

+ =

−

5 5 6 5

5 5 5 6

5 5

5 52 3

1

2 3

1

2 3

1 3

5

( )

( )

+ ⋅ − =

−

− =

−

− =

−

− =

−⋅

−

=

+

−1

2 3 3

1 5

32 9

3 5

37

2

37

2

3

1

7

2

21

( )


− −

−=

−

=

−

= −

1

16

1

21

16

1 8

161

16

7

161

16

77

16

16

1 7⋅ = −

Mergulhando Fundo

01 E

x = 0,00375 · 10–6

e y = 22,5 · 10–8

x = 3,75 · 10–9 e y = 225 · 10–9

y

x=

⋅

⋅

=

−

−

225 10

3 75 1060

9

9,, logo, y = 60x

02 2x = 23y + 3 32y = 3x – 9

x = 3y + 3 2y = x –9

x y

y x

= +

= −

3 3

2 9

Utilizando o método da substituição, tem-se:

2y = 3y + 3 – 9 x = 3 · 6 + 3 –y = –6 · (–1) x = 21 y = 6 x + y = 21 + 6 = 27




LIVRO 1 | ÁLGEBRA

03 a) − = − ⋅ = − ⋅ = − ⋅ = − = −2 744 2 7 2 7 2 7 14 143 3 33 33

3

3 1. ( ) ( ) ( )

b)343

1 000

7

10

7

10

7

10

7

10

7

10 03

3

33

3

3

3

31

. ,= =

=

=

= = 77

c) − − = − − = − − = − = +4 913 17 17 17 173 33

3

3 1. ( ) ( ) ( )

d) − =

−

=

− +

= −

∴

∴ −

0 000064 64

1 000 000

2

10

2

10

2

10

3 3

6

6

3

6

3,. .

( )

= −

= − = −

6

32

2

10

4

100 0 04,

01 M x x x

M x x x

M x x

= + + >

= ( ) + ⋅ ⋅ + = +( )

= + = +

4 4 0

2 2 2 2

2 2

22

2

,

02 B

x x x x

x x x

x x

x x

x x

x x

+ + − < <

+ ( ) − ⋅ ⋅ +

+ −( )+ −

+ − +

− + = +

1 2 0 1

2 1 1

1

1

1

1

22

2

,

*

( )

11

* Como 0 < x < 1

x x x− = − −( ) = − +1 1 1

03 a) 7 7 7 71 1= =, pois

b) − ∃ − ∉356 3564R R

+, pois

c) − = − = − − = −127 13 13 13 1273

3

3

3

, pois

d) 0 04 4

100

2

10

2

10

4

100

2

, ,= =

=pois

04 a b2 23 2 23 3 3 32 8 6 2 100 2 10 2 8 2+ − = + − = − = − = =

05 (F), pois x x2=

(V)(V)

(V)

(F), 49 7=

(V) (F) 8 2 23 33

= =

(F) − = − ∉3 92R

01 a) R +

b) R +

c) R *– , a ∉R

d) Existe

e) a se e a se3 30 0 0 0≥ ≥ < <, ,a a

f) a b e= ⇔ = ≥b a b2 0

g) a b3 3= ⇔ =b a

02 a) 1 764 2 3 7 2 3 7 2 3 7 422 2 22

2

2

2

2

2 1 1 1. = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ =

b) 15 876 2 3 7 2 3 7 2 3 7 1262 4 22

2

4

2

2

2 1 2 1. = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ =

c) 5 856 400 2 5 11 2 5 11 2 5 11 2 4204 2 44

2

2

2

4

2 2 1 2. . .= ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ =


03 −÷ +

÷

÷

− −

− − −( )

( )( ( ) ) ( ) (

10

1010 10 10

4 9

2 27

1 9 2 3 4 3 ))4 1

36

54

18 12 12

1

10

1010 10 10

10

−÷

÷

÷ ∴

∴ −

−

− −

− 88 18 12 12

12 12 12

10 10 10

1 10 10 10 10 112

÷ ÷{ } ÷ ∴

∴ − ÷{ } ÷ ⇒ − ÷ = −

− −

− −

04 I. P = 32 · 7 · 1050 · 231 · 540 · 26

P = 32 · 7 · 1050 · 237 · 537 · 53

P = 32 · 53 · 7 · 1050 · (2 · 5)37

P = 32 · 53 · 7 · 1050 · 1037

P = 32 · 53 · 7 · 1087

Logo, 87 zeros.

II. A2 – B2 = (A + B) · (A – B)

3 3

2

3 3

2

3 3

2

3 3

2

3 3 3 10

x x x x x x x x

x x

++

−

⋅

+−

+

⇒

⋅ = =

− − − −

−

III. a) 3n + 2 · 2n + 3 = 2.592 = 25 · 34

n + 3 = 5 ou n + 2 = 4, logo n = 2

b) 3n · (1 + 3 + 9 + 27) = 1.080

3n · 40 = 1.080 ⇒ 3n = 1 080

40

. ⇒ 3n = 27 = 33, n = 3

Radiciação Aula 1Capítulo 2


04 a) 676 2 13 2 13 2 13 262 22

2

2

2 1 1= ⋅ = ⋅ = ⋅ =

b) 0 216 216

1 000

6

10

6

10

6

10

6

103 3

3

33

3

3

3

31

,.

= = =

=

=

∴∴

∴ =6

10 0 6,

c) 12 167 23 23 23 23

3 333

3 1

. = = = =

d) 3 136 2 2 2 7 56 3 136 566 7 3 2 2. ( ) .= ⋅ = ⋅ = ∴ =

e) − ∃625 R

f) − = −

= −

= −

= −

729

2 197

9

13

9

13

9

13

9

133

3

3

3

31

.

05 B

16 4 4 162= ⇔ =




LIVRO 1 | ÁLGEBRA

01 a) 8 82

3 23=

b) 6 61

2 1=

c) a a

3

5 35=

d) x x1

4 14=

e) 9 1

9

1

9

1

21

21

−

= =

f) 17 1

17

1

17

1

17

3

53

5

3

535

−

= = =

g) ( )

( ) ( )a b

a b a b

2 31

6

2 31

62 3 16

1 1−

= =

h) ( ) ( ) ( ) ( ),xy z xy z xy z xy z xy z2 0 4 2

4

10 2

2

5 2 252

2

5 1

1

−

− −

−= = = = ∴

∴xx y z2 4 2

5

i) a

a a

b

bb

−

= =

5

5 5

1 1

02 a) 4 0 5 0 25 8 4 0 0625 0 5 8

0 25 0 5 64 0 25 0 5

42

3 23

3

⋅ + + ⇒ ⋅ + +

+ + ⇒ + +

( , ) , ( , ) ,

, , , , 44 4 75⇒

,

b) 100 81 16 1

10081 16

1

1003 16

1

0 51

4 0 25

0 5

4

25

100

25

25

1

4

−

÷

÷

+ − ⇒ + −

+ − ⇒

, ,

,

1103 16

1

103 2

1

101

1 10

10

11

10

4+ −

+ − ⇒ + =+

=

c) ( )

( ) ( ) ( )

( )

, ,

, ,

− + −

− + −

− + −

− −

− −

−

1 36 256

1 6 2

1 6

3

7 0 5 0 25

3

7 2 0 5 8 0 25

3

7 1 221

1

1

6

1

4

12 2 3

12

13

12

2−

− + − =− + −

= −

03 K

K

= + −

+

−

= ( ) + ( )

− −

36 64 1

121

1

4625

6 4

1

2

1

3

0 5 1

0 25

21

2 3

,

,

11

3

2 0 5 1

4 0 25

1 1

1

11

1

45

6 4

1

11

−

+

− ( )

= + −

− −,

,

K

+

−= + − + −= −

− −1 1

11

4 56 4 11 4 5

2

K

K

01 C

8 2 2 83 3= =, pois

02 ( )

( ) ( )

,

2 5 5

2 5 2 5 2 5 5 2

2 5 5 5 2 5 2

2

2

− −

− = − = − − = −

− − ⇒ − − = −Logo

03 a) a ∈ R + e b ∈ R

+

b) (a, b) ∈ R

04 M x x x x

x x

= − + = − ⋅ ⋅ + ∴

∴ − = −

4 4 1 2 2 2 1 1

2 1 2 1

2 2 2

2

( ) ( )

( )

a) M x M= − = ⋅ − = − =

−

= − =2 1 2 1

81

1

41

1 4

4

3

4

3

4

b) M x= − = ⋅ − = − = =2 1 2 12

1 1 1 0 0

c) M x= − = ⋅ − = − = =2 1 2 19 1 38 1 37 37

05 K x x x x x= + + + = + = +3 23 333 3 1 1 1( )

∴ para x = 1,996, tem-se 1,996 + 1 = 2,996.

06 0 < b < a e ab > 2

a)a ab b

a ab b

a b

a b

a b

a b

2 2

2 2

2

2 2

2

2

+ +

+ +

=+

+

=

+

+

( )

( ) ( )

como 0 < b < a a b a b

a b

a b a b+ = + =

+

+

=

+

( )

( )2

1

04 a) 5 51

2=

b) 36 3651

5=

c) 4 4383

8=

d) x x373

7=

e)

1

2

1

2

9 9

2 =

f) 2

1

3

2

1

3

2 1

33

5

3

5

3

5

=

= ⋅

−

05 a) 49 49 7 751

5 21

5

2

5= = ( ) =

b) − = −7 731

3( )

c) 8 8 2 241

4 31

4

3

4= = ( ) =

d) a a a1520

15

20

5

53

4= =

÷

÷






LIVRO 1 | ÁLGEBRA

b) a ab b

a a b ab b

a b

a b

a b

a b

a b

a b

2 2

3 2 2 33

2

33

2

3 3

− +

+ + +

=−

+

=

−

+

=−

+

( )

( )

Observação: |a – b| = a – b, pois 0 < b < a.

c) a b ab

a b

ab

ab ab

ab

ab ab

2 2

2 2

24 4

4

2

2 2

2

2 2

− +

−

=−

+ ⋅ −

=

−

+ ⋅ −

=

=

( )

( ) ( ) ( ) ( )

(aab

ab ab ab

−

+ ⋅ −

=

+

2

2 2

1

2

)

( ) ( )

Observação: |ab – 2| = ab – 2, pois ab > 2.

01 a) a a a a2020

2 10 5 2 23 9= = = = =( )

b) a a a15315

3 5 3= = =

c) a a a a75375

3 25 5 5 53 243= = = = =( )

02 a) ( ) ( )− = − = −5 5 51

5 15 5

b) ( , ) ( , )0 4 0 43

4 34=

c) −

= −

1

9

1

9

3

73

7

d) ( ) ( )abx abx abx1

2 1= =

03 a) 8 1

25 0 017

9

8 2 5 1

1

3

1

20 3

1

3 21

2+

⋅

⋅ ⇒ + ⋅

−− −

, ( ) ( )

⋅ ⇒

⇒ + ⋅ ⋅ ⇒ + ⋅ = ⋅ =

9

8

2 5 1 9

8 2 5 9

8 7 9

8

63

81 1

{ } { }

b) 2 2

50 3444 9

3

21

20 1 53

3

−− −

+

÷ ( ) +

⋅

⇒

⇒

, ... ,

11

2

25

41 9

8

27

2 25

41 9

8

2

1 5

3

1

2+

÷ +

⋅ ⇒

+

÷ +

⋅

,

77

27

41 3

8

27

27

4

1

4

8

27

2

4

1

2

2

2

⇒

⇒

÷ + ⋅ ⇒ ⋅ ⋅ ⇒ =

÷

÷[ ]

04 a) 10 79 6 4 4 489

10 79 6 2 67 10 79 8

34

34 2 34

⋅ + +

⋅ ⇒

⋅ + +( ) ⋅ ⇒ ⋅ +( ) ⋅

( . )

( ) 667

10 79 2 67 10 81 67 10 3 67 2 0104 4

⇒⋅ +( ) ⋅ ⇒ ⋅ ⋅ ⇒ ⋅ ⋅ = .

b)2 2

10

2 2 2

10

2 10

10

2 2 2 512

28 30

3

27 3

3

27

3

27327

3 9

+=

⋅ +=

⋅⇒

= = =

( )

01 a) a ∈ R +

b) n ∈ R e n ≥ 2

02 a) x x x x777

7 1= = =

b)

16

9

16

9

16

9

9

16

5

5

5

51

=

=

=

− −

−

c) m m m

m

−

−

−

= = =66

6

6 1 1

d)

m n

p

m n

p

m n

p

p

m m p mp

x

x

x

x+

=

+

=

+

∴

∴+ +

− −

−3 3

3

3

3 2 23 3 ++ p3

03 x y x y x y x ym mm mm mm

m

m

m

m3 4 3 43 4

3 4⋅ = ⋅ = ⋅ = ⋅

04 a) 2 187 32 2 187 32

3 3 2 3 9 2 12 9

25 5 25

5 25 105 5 2 5

. ( ) . ( )⋅ = ⋅ =

⋅ ⋅ = ⋅ ⋅ =

b) ( . ) ( )

.

343 2 197 7 3 7 3

7 3 3 7 3 3 3 969

23 3 7 23 6 143

6 12 23 2 4 23

⋅ = ⋅ = ⋅ =

⋅ ⋅ = ⋅ ⋅ = 993

05 a) 2 2 2 2 2 2 4 2204 208 16 48 2 4 48 4= = ⋅ = ⋅ = ⋅

÷÷

b) 4 096 2 2 845 12 440 4 310 10. = = =÷÷

c) 81 3 352 4 420 4 5= =

÷÷

d) 256 2 2 165 8 210 2 45 5= = =

÷÷

06 a)3 3

3

3

33 3 9

1421 1021

1021

2421

1021

14 721 7 23 3⋅

= = = =÷÷

b) x x y

xy

x y

x yx y y x

5 7

5

6 7

5 5

12 6⋅ ⋅

= = ⋅ = ⋅

− −

− −

( )

01 b

ab

a a b ab

ab

− ÷

÷

= = =

=

=

2

2

1

4 2

2

4

2 24

2 2

4 2

16

1

16

1

16

1

4

1

4

1

2 (( )ab − 1

02 a)x x y

y

x x y

y

x x x y x y

y

x x y

10 4

93

10 43

93

9 33

3

3

⋅ +=

⋅ +

=

⋅ ⋅ + ⋅ +

=⋅ +

( ) ( ) ( ) ( )

( )

yyx x y

33

⋅ ⋅ +( )

b)a

x

a

x

a a

x x

a a

x x

a

x

a

x

n

nn

nn

nn

n nn

n nn

nn

nn

3

52

32

52

22

42

2

2 2= =

⋅

⋅= =

n

n2

c) x x x x x x x x

x x

3 2 2 210 25 10 25 5

5

− + = ⋅ − + = ⋅ − ∴

∴ − ⋅

( ) ( )

( )

d) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

a b a b a b a b a b

a b a b a b a b

2 2

2

− ⋅ + = + ⋅ − ⋅ + ∴

∴ + ⋅ − = + ⋅ −

Mergulhando Fundo


Radicais Aula 1Capítulo 3





LIVRO 1 | ÁLGEBRA

01 a) 7 7 768 6 28 2 34= =

÷÷

b) a b a b a b ab3 69 3 39 3 6 39 3 3 23 23⋅ = ⋅ = ⋅ =

÷÷ ÷÷

c) x

a b

x

ab

x

ab

4

2 610

2

3

2 2

10 2

2

35=

=÷

÷

d) 32 2 2

2

1510 5 510 5 10 5 15 5 3

3

⋅ − = ⋅ − = ⋅ − ∴

∴ ⋅ −

÷÷÷

÷

( ) ( ) ( )

( )

m n m n m n

m n

e) ( ) ( )a b a b a bou

a b+=

+=

+ +2

612

212

612

6

6

2 2 2 8

f)

( )p q

a b

p q

a a b

+=

+

⋅

9

129

39

02 a) 12 2 3 2 32= ⋅ =

b) 75 3 5 5 32= ⋅ =

c) 48 2 3 2 34 44 4= ⋅ =

d) 180 2 3 5 2 3 5 6 52 2= ⋅ ⋅ = ⋅ ⋅ =

e) 243 3 3 3 3 34 54 44 4= = ⋅ =

f) 56 2 7 2 73 33 3= ⋅ =

03 a) 25 52 2x y x y= ⋅

b) 16 42mx x m= ⋅

c) 64 46 123 2 4a m a m=

d) 75 5 3 5 33 3 3 2 2 2 2a b c a a b b c c abc abc⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =

e)3

2

3

2 2

3

2 2

3

2

4 2 8

6 64

4 2 4 4

4 2 4 24

2 2

2 24

2

m n

p

m n n

p p

n

p

m

p

n

p

⋅

= ⋅ ⋅ ⋅

⋅ ⋅ ⋅

= ⋅

⋅ ⇒

⋅ mm

p

n

p

m

p2

3

2 2

2 2

4 2

2

= ⋅

÷

÷

f)250 5 2 5 10 5 10

3 4

3

2 4 2 2x y a x xy a xy xa xy xa=

⋅

⋅ ⋅

= =

04 a) a ab b a b a b2 2 22+ + = + = +( )

b) 81 18 9 92 2+ + = + = +x x x x( )

c) x xy y x y x y2 24 2 24 22− + = − = −÷÷ ( )

d) a a a a a a a a a a3 2 2 22 2 1 1 1− + = ⋅ − + = ⋅ − = − ⋅( ) ( ) ( )

05 a) 2 8 2 2 2 2 2 2

4 2

3 4 2 2 4 2

2 2

m m n m m m n m m n m

m n m

= ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ ∴

∴

03 a) 5 3 5 3 753 2 6 2 6 2x y x y x y= ⋅ ⋅ =

b) 3 3 3 3 813 3 3 93 103a a a a a= ⋅ ⋅ =

c) 1 1

4 3

3

4 3

3

4 3

9

16 3

3

16

2 3

3−

⋅ = ⋅ =

⋅ = ⋅ =

÷

÷

y y y y y

d) 1

1

9 4

10

9 4

10

9 4

1 000

729 4

4 0

2 23 2 23

3

6 23

6 23

a a a a a a

a a

= =

⋅ ⇒

⋅ ⋅ =. . 000

729

8

3 a

04 Fx

a

x

x

aF

x

a

x

x x

a

Fa

a

= ⋅ ⋅ ⋅ ⇒ = ⋅ ⋅ ⋅⋅

⋅

= =

⋅ ⋅

= =

⋅

3 4

3 2

3 4

3 2

4

2

4

2 2 2

4

2 4

4

2 2

2 3

4 2

== =4

41

05 a) 5 2 3

5 2 3

23 53 55

6 15 25

,

,

ou

ou

m.m.c. (6, 15, 25) = 150

5 2 3

5 2 3 5

25150 10150 6150

25150 10150 6150 23

ou

o

⇒

> > log , . b) 3 2 5 4

6 20

6 20 216 400

5 4

3 3

3

36 26 6 6

3 3

⋅ ⋅

⇒

⋅

ou

ou

ou ou

Logo, .

c) 10 7

10 7

10

1 000 117 649

7

3

4

3

2

3

4

6

4

34

4 4

3

2

ou

ou

Logo

;

. .

,

06 E ab

c

ab

c

ab

c

n

n

n

n

=

=

=

=

⋅ −

−

3 3

35 4

1

2

( )

33 3

3

20

1

2

40 64 000

= −

−

∴

∴ =[ ] .

07 a) a b c a b b c a b b c

a b bc

n n n n n n

n

4 4 14 44 44 4 4 4

4

⋅ ⋅ = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ∴

∴ ⋅ ⋅

+

( ) ( )

b) 32 343 243 2 7 3 2 2 7 3 3

2 2 7 3 3 168

53 10 3 53 93 33 3 23

3 3 23

⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ = ⋅⋅ 183

08 a) ( ) ( )16 9

5

2 3

5

2 3

5

2 2 3 3

5 5

2 3

5

73

4 2 5

73

20 10

73

18 23 93

63

6 3

⋅

=

⋅

=

⋅

=

⋅ ⋅ ⋅

⋅

=

⋅ ⋅⋅

=

12

5 5

1 728

25

12

5

3

2 33

.

b) ( , )0 08

8

100

2

25

25

2

5

2

5

2 2

12

3

3 3 3 6

3

3

−− −

=

=

=

= ∴

∴ = 55

2 2

c) 32 2

2 2

7 165 55 5 25 15 15

3 2 153

2

5

x y x x y y

x y x y x

y

x

y

− − −

− −

= ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ =

d)2 3

52 3 5 360

3 2

5 3 25 5n n

n

n n n n⋅

= ⋅ ⋅ =−

( ) ( ) ( )

Não é possível extrair o radical.





LIVRO 1 | ÁLGEBRA

b) 2 32 2 2 2 22

3

12

4 55

2

3

5 10 2

4 55

2 2

3

2

45

a

b

b

a c

a

b

b b

a c

a b

b c

b

c= ⋅

⋅ ⋅

⋅

=

⋅ ⋅

⋅

⋅ ∴

∴

44 42 2

3

2

45

2 2

45

a b

b c

b

c

a

bc

b

c=

c)m n m n

m b

m n m n

m b

m n

m bm n

+

⋅

−( )

+( )=

+

⋅

−( )

+( )∴

∴+

⋅ +

⋅ −

2 2

2

2

63

23

2

2

23

( )( )

d) a b

a

a

a b

a b

a

a a

a b a b

a b

a

aa a

−

−( )=

− ⋅ ⋅

− ⋅ −

∴

∴

−

⋅

⋅

2

8

2

8

2

8

13

65

10 35

55

2

( ) ( )

335

5

3

5

2

8

( ) ( )a b a b

a a

a b− ⋅ −

=

−

01 E x x

x x

x x x x

x x

x

=+ ⋅ +

− +

=+ ⋅ − + ⋅ +

− +

∴

∴ + =

( ) ( ) ( ) ( ) ( )

( )

( )

3

2

2

2

2

1 1

1

1 1 1

1

1 xx + 1

a) |0,47 + 1| = 1,47b) |–349 + 1| = 348

c) 3 444 1 4 44 40

9, ... , ...+ = =

02 y y y y y y y y y y

1

4

3

4

1

2

1

4 2

1 1

4

3

4

1

2

1

4 21

⋅ + + +

⇒ ⋅ + + +

−

⇒

⇒ + + + ⇒ + + + ⇒

⇒ + + +

y y y y13

4

1

2

9

4 4 1 43

4 41

2 49

4

4 3 2

2 2 2 2

2 2 2 2

( ) ( ) ( ) ( )99 16 8 4 512 540⇒ + + + =

03 a b

ca

b

ca

b

ca

b

c

ac b

c

a b

cac b a b

ac a

+

=

⇒ + = ⇒

⇒ +

= ⇒ + =

⇒ =

2 2

2

22

222 1

b b c ab a

a− ⇒ =

⋅ −( )

04 3 3

3 3

3 3

3 3

3

3

3

33

3

220 415

330 6

10 415

10 6

415

6

830

530

330

3 3

⋅

⋅

=

⋅

⋅

= = = ∴

∴ ÷660 3 20 3÷

=

05 9 2

9 2

3 2

3 2

9

34 0 333 0 111

123

9

1

9

2121

27

6 27

ou

ou

ou

ou

, ... , ...

( )

334

06 M b

a

a b

b bM

b

a

a b

b

M b

=−

− +

⇒ =

−( )⋅

−( )∴

∴ = = =

1

3

9

2 1

1

3

3

1

676 26

2

2

01 a) 2 2 2 233 33 433 49⋅ = =

b) 5 5 5 5 5 5 54 6105 2010 6105 26105 26 250 2 1325⋅ = ⋅ = = =

÷÷

c) 2 2

2

2

22 2

4530 3430

20303

7930

20303 59303 5990⋅

= = =

02 E x x x x x= − + − = −( ) = −

− =

−

=

3 23 333 3 1 1 1

40

11 1 40 11

11

29

11

03 12 1

21

2 1

4

4 2 1

4

1

8 4 2

2 2

8 4

4

4 8 4

4

4 2

+ − +( )

( )= +

− +=

+ − +∴

∴ +( )

x x

x

x x

x

x x x

x

x

22

1

2

1

22 2

4

2

2 4

2x

x

x

x

x( )=

+

=

+

04 En n

nn n

n

nn

=

⋅ − ⋅

=

⋅ − ⋅

=

⋅ ⋅ −

=

600

25 25 5 5

600

5 5 5 5

600

5 5 5 1

6

2 2 2 2 4 2 2

2 2 2( )

000

5 600

1

5

1

5

1

252 2 2nn

nn

⋅

= = =

01 a) 8 18 32 50 2 4 2 9 16 2 25 2

2 2 3 2 4 2 5 2 14 2

+ + + = ⋅ + ⋅ + ⋅ + ⋅ ⇒

⇒ + + + =

b) 12 48 75 4 3 16 3 25 3

2 3 4 3 5 3 3

+ − = ⋅ + ⋅ − ⋅ ⇒

⇒ + − =

c) 8 3

42

3

168

3

42

3

16

8 3

2

2 3

4

7 3

22

− = ⋅ − ⋅ = − =

d) 75 48 3 108 25 3 48 3 36 3

5 3 48 3 6 3 37 3

− + = ⋅ − + ⋅ ⇒

⇒ − + = −

e) 2 16 54 128 2 8 2 27 2 64 2

2 2 2 3 2 4 2 10 2

3 3 3 3 3 3 3 3

3 3 3 3 3

+ + + = + ⋅ + ⋅ + ⋅ ⇒

⇒ + + + =

f) 162 1 250 2 592 81 2 625 2 1 296 2

3 2 5 2 6 2 8 2

4 4 4 4 4 4

4 4 4 4

− − = ⋅ − ⋅ − ⋅ ⇒

⇒ − − = −

. . .

Operações com radicais I


Mergulhando Fundo

Aula 1Capítulo 4





LIVRO 1 | ÁLGEBRA

01 a) a b ab6 6 6⋅ =

b) 144 6 2 3 2 3 2 3 2 3 2 33 3 4 23 3 3 33 3÷ = ⋅ ÷ ⋅ = ⋅ = ⋅ =

c) 2 3 65 25 5 45x x x x⋅ ⋅ =

d) ( )8 4 2 32 2 16 234 34 4 64 4 54 4a a a a a a a a⋅ ÷ = ÷ = =

e) ( . ) .

.

162 16 000 12 15 216 000 15

14 400 120

⋅ ÷ ÷ ⇒ ÷ ∴

∴ =

f) 8 18 2 3 4 3 124 2 2a a a a a⋅ = ⋅ ⋅ = ⋅ ⋅ =

g) 2 14

22 7

xy

yx=

h) a a a a a a6 3 3 3 3 31 1 1 1 1 1− ÷ + = + ⋅ − ÷ + = −( ) ( )

i) 2 2 2 2 214 34 1 34 24x x x x− − − − +

÷ = = =

02 a) x x x x x215 310 430 930 1330⋅ = ⋅ =

b) a a a a a a4 12 312 112 212 6÷ = ÷ = =

c)8

4

8

4

64

64

2 524

216

2 5 248

2 348

448 10

348 6

448x y

xy

x y

xy

x y

x yxy= = =

( )

( )

d) m m

m

m m

m

m

mm

215 1720

1130

860 5160

2260

5960

2260

3760⋅

=

⋅

= =

e) ( )

(

9 32 6 8 27 16

2 2

4 28 4 5 312 5 3 26

15 12 6

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ÷ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅

a b c a b c a b c

a b ⋅⋅ ⋅ ⋅ ⋅ ⋅ ÷ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ÷

c a b c a b c

a b c

324 6 8 10 624 16 20 12 824

21 20 16 924

2 2

2 2

)

22 2 22 32

16 20 12 824 5 424

424

⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ∴

∴

a b c b cb c

03 a) 2 4 3 6 4 6 6 9 2 4 7 6 6 93 3 3 3 3 3 3− − + ⇒ − +

b) 6 12 4 18 12 12 4 6 8 2

6 4 3 4 9 2 12 4 3 4 6 8 2

12 3 12 2 24 3 4 6

− − + + =

= ⋅ − ⋅ − ⋅ + + =

= − − + + 88 2

12 3 4 2 4 6

=

= − − +

04 A

B

= − = − =

= − = − =

+

=

9 2 10 81 40 41

7 2 3 49 12 37

41

37 4

4

2 2

2 2

2 010

( )

( ).

11

41 1 1

2 010

2 010

= =

.

.

05 a)ab b

a ab

a ab

ab b

b a b

a a b

a a b

b a b

a b

+

−

⋅+

−

=⋅ +

⋅ −

⋅ +

⋅ −

∴

∴+

2

2

2

2

2

( )

( )

( )

( )

( )

(( )a b

a b

a b−

=+

−2

b) x y x y x y

x y x y x y

x y x y

+ ⋅ + ÷ +

+ ⋅ + ÷ +

+ = +

3 4 12

412 312 12

612

( ) ( ) ( )

( )

06 A B

A B

A A B

= − ⋅ + ⋅ = ⋅ − ⋅

= − + = − =

=−

5 2 3 5 2 7 12 2 5 2 2 3 3 2

5 2 15 2 84 2 10 2 9 2 2

74 27

2

33

73 2

73 2= =

07 x x x= − ⇒ = − ⇒ = − =2 27

91

2 3

31 2 1 1

6

4

612

612

08 x x= − = = =36 27 9 9

09 a) 3 5 3 3 2 5 3 3 5− − + =

b) 7 15 2 4 8 5 3 4 5 5 20 2 20 5

105 2 32 5 12 5 100 2 20 5 5 2

⋅ − ⋅ + ⋅ − ⋅ + ⇒

⇒ − + − + =

c) 15 25 40 15 5 2 10− + = − +

d)3

25

2

53

1

310 6

150

56

5 6

56 0

⋅ ⋅ ⋅ ⋅ − ⇒ − ∴

∴ − =

10 a) ( ) ( )15 3 15 3 1− ÷ − =

b) x + y – y = x

c)( )

( ) ( )

a

a a a

−

+ ⋅ −

=

+

5

5 5

1

5

d) 3

48

3

42

3

233 m m m= ⋅ ⋅ =

Mergulhando Fundo

01 A = 49 – 63 = –14B = 25 – 48 = –23

− −− −

=

−

= −14 23

23 14

37

37 1

2 010 2 010

2 010

( ) ( )

. .

. == 1

02 2 243 1 2 27 32 9 3 2 2 3 3 6

18 3 2 6 3 6

2 24 3 4

⋅ − + ⋅ + =

= ⋅ − + ⋅ + =

= − + + ⇒

⇒ = +

( ) ( )

p

A

A

A

= − ⋅ + = + − −

= + − −

= +

( ) ( )3 1 3 3 3 3 3 3 3

81 27 3 3 3 3

24 3 78

5 3 8 5 3

03 266

1914= cm

04 a) 18 9

12 9

18

12

9

9

3

29

2

3

3 36

26

36a a

a

a a

aa a

−

−

= ⋅

⋅ −

−

= ⋅ ⋅ −

( )

( )( )





LIVRO 1 | ÁLGEBRA

01 Substituindo x por –1, tem-se:

−( ) + − −( )

−( ) + + −( )=

+

−

1 2 2 1

1 2 2 1

3 2

3 2

2

2

Racionalizando o denominador, tem-se:

=

+ +( )− +( )

=+ + +

+ − −

∴

∴+ +

−

=+

−

3 2 3 2

3 2 3 2

3 2 3 2 3 2

3 2 3 2 3 2

3 4 3 4

3 4

7 4 3

1

2 2

2 2

== − −

7 4 3

02 a)x y

x y

x y

x y

x y x y

x yx y

−

−

⋅

+

+

=

− ⋅ +

−

= +

( )

( )

( ) ( )

( )

b)a a

a a

a a

a a

a a a

a a

a a

+ − −

+ + −

⋅+ − −

+ − −

=

=+ − − + −

+ − +

=−

1 1

1 1

1 1

1 1

1 2 1 1

1 1

2 22 2−−

= − −1

212a a

c) a mm a

m am a

a m m am a

m a m a

2 2

2

1

−

−

⋅+

+

=− ⋅ +

−

∴

∴ − ⋅ + = − −

( )( )

( )

( )

d) a a

a a

a a

a a

a a a a

a a

a

+ −

− −

⋅+ −

+ −

=+ − + −

− −

∴

∴

2

2

2

2

2 2 2

2 2

2

1

1

1

1

2 1 1

1

2

( )

( ) ( )

++ − −

+ +

=+ − −

∴

∴ + − = + − −

2 1 1

2 2 1

2 2 1 1

1

2 2 1 2 2 1 1

2

2

2 2

2 2 2 2

a a

a a

a a a

a a a a a a

03 18 7 18 7 8 7 8 7 2 8 4 2+

+

−

= − + + = =

04 am an

m n mn

a m n m n

m mn n mn

a mn m n

m n

2 2

2 2 2 2

−

− +

=⋅ + ⋅ −

− + +

⋅ ⋅ −

+

( )

( ) ( )

( ) ( )

( )) ( ) . ( ) . .= ⋅ − = + − = ⋅ =a m n n n2 011 1 2 011 1 2 011

05 E

E

E

E

=

+

+

−

− ∴

∴ =−

−

++

−

− ∴

∴ = − + − − − ∴

∴ = −

1

2 3

1

2 32 2

2 3

1

2 3

1

2 2

2 3 2 3 2 2

4 2

f)10

2 3

10

4 3

10

12

10 12

12 12

10 2 3

12

10 2 2

4

34

4 34

2 34

44

4 2

=

⋅

= =

⋅

⋅

∴

∴

⋅ ⋅

=

⋅ ⋅( ) ⋅⋅

∴

∴

⋅ ⋅ ⋅

=

⋅

=

⋅

3

12

10 2 4 27

12

20 108

12

5 108

3

34

4 4 4

g)a

a b

a ab

a b ab

a ab

a b

a ab

ab

ab

b34

34

34 34

34

4 44

34 34

=

⋅

⋅

=

⋅

= =

h) 5

5

5

5

5 5

5 5

5 5

5

5 5

53 125

3 6

56

6 56

56

66

566

= =

⋅

⋅

= = = .

02 a) 6

2 2

2 2

2 2

6 2 2

4 2

6 2 2

23 2 2

+

⋅−

−

=⋅ −

−

=⋅ −

= ⋅ −( )

( )

( ) ( )( )

b)3

3 6

3 6

3 6

3 3 6

9 6

3 3 6

3 3 6

−

⋅+

+

=⋅ +

−

=⋅ −

= +( )

( )

( ) ( )

c)3 2 2

3 2 2

3 2 2

3 2 2

3 2 2

1

2+

−

⋅

+

−

=+( )

( )

( )

d) 3 2

1 2

1 2

1 2

3 3 2 1 2 2

1 2

5 4 2

15 4 2

−

+

⋅−

−

=− − +

−

∴

∴−

−

= − +

( )

( )

e)3 2

7 2 6 3

7 2 6 3

7 2 6 3

42 18 6

98 108

42 18 6

10

42 18 6

10

−

⋅+

+

=+

−

∴

∴+

−

=− −

( )

( )

f)17

3 5 2 7

3 5 2 7

3 5 2 7

17 3 5 2 7

45 28

17 3 5 2 7

173 5 2 7

+

⋅−

−

=⋅ −

−

∴

∴⋅ −

= −

( )

( )

( )

( )

03 a)2

2

3

3

2 2

2

3 3

32 3+ = + = +

b)2

3

3

2

2 3

3

6

2

4 3 3 6

6+ = + =

+

c)1

2

2

4

3

8

4

16

1

21

3

2 21

2

2

3 2

4

2

1

2 2 3 2 8

4

5 2 8

4

+ + + = + + + =

= + + =+ +

=+

04 A =

+

⋅−

−

=⋅ −

−

= −3

5 2

5 2

5 2

3 5 2

5 25 2

( )

( )

( )

B =

+

⋅−

−

= ⋅ −

−

= −13

3 5 4 2

3 5 4 2

3 5 4 2

13 3 5 4 2

45 323 5 4 2

( )

( )

( )

C = − − − = ⋅ − − + ∴

∴

4 5 2 3 5 4 2 4 5 4 2 3 5 4 25

( ) ( )

05 4 3

4 3

4 3

4 3

4 3 4 3

4 34 3

+

+

⋅+

+

=+ ⋅ +

+

= +( ) ( )





LIVRO 1 | ÁLGEBRA

01 a)1

3 2 6

3 2 6

3 2 6

3 2 6

3 2 2 6 6

3 2 6

2 6 1

2 6 1

2 6

( )

( )

( )

( )

(

+ −

+ +

+ +

=+

+ + −

∴+ +

−

+

⋅+

∴

⋅

++

∴

∴+ + + + +

−

+ + + + +∴

+ + +

∴

∴

∴

1

2 18 3 2 12 2 12 6

24 1

6 2 3 4 3 2 12 6

23

7 2 5 3 6 12

23

)

06 a) 6 2 5 1 5 1 5 5 12− = − = − = −( )

b) a b a b a b a b

a a b b

a b b

2 2 2 2 2 2 2

2 2 2 2

2 2

2 4

2

2

2

2

− − = − ⋅ − ∴

∴+ −

− ∴

∴ − −

( )

c) 7 48 2 3 2 3 2 32+ = + = + = +( )

d) 23 45 23 22

2

23 22

2

45

2

1

2+ =

++

−= +

07 a) 3 2

3 2

3 2

3 2

3 2

1

3 2 3 2

2−

+

⋅−

−

=−

∴

∴ − = −

( )

b) 7 2 6 6 2 5 1 6 1 5

1 6 1 5 1 6 5 1 6 5

2 2

+ + − ⇒ + + − ⇒

⇒ + + − ⇒ + + − ⇒ +

( ) ( )

08 a) 10 96 10 2

2

10 2

26 2± =

+±

−= ±

b) 57 12 15 57 2 160 57 33

20

57 33

2

45 12

− = − =+

−−

∴

∴ −

.

c) 3 1

210

7

210

7

2

3

22

2

51− = − =

−

⇒ −

d) 17 12 2 17 288 17 1

2

17 1

2

9 8 3 8

− = − ⇒+

−−

∴

∴ − ⇒ −

09 m m m m m m

m

+ + =+ + −

++ − +

∴

∴ +

3 12 3 3

2

3 3

2

3

10 a

ba b=

+

+

⋅−

−

=−

−

= = =5 1

3 1

5 1

3 1

5 1

3 1

4

22 2

b) 11

6 3 2

11

6 3 2

6 3 2

6 3 2

11 6 5 6 3 2

19

4 4

4

4

4

− −

=

− +

⋅+ +

+ +

=

=⋅ − ⋅ + +

( )

( )

( )

( ) ( )

02 a) 3

4 2 3

4 2 3

4 2 3

3 4 2 3

4

6 2 3

43 2 3

2

20

2 3

2 3

2 3

−

⋅+

+

=⋅ +

=+

=

=⋅ +

=

+

⋅−

−

( ) ( )

( ) ( )

( ))( )= ⋅ −20 2 3

Logo, 20

2 3+

b)5 2

7 3

5 2

7 3

3

41

−

+

⋅+

−

= <

a < b. Logo,7 3

5 2

−

+

03 I. 5 2

5 2

5 2

5 25 2 5 2

5 2 5 2

5 2 5 2

5 2 5 2

+

−

+−

+

∴

+ ⋅ +

− ⋅ +

+− ⋅ −

+ ⋅ −

( ) ( )

( ) ( )

( ) ( )

( ) ( ))

( ) ( )

∴

∴+

−

+−

−

∴

∴ + + − =

5 2

5 4

5 2

5 4

5 2 5 2 2 5

2 2

II. E x

x

x

xE

E E

E

=

−

−−

⇒ =

−

−

−

⇒

⇒ = − ⇒ = ⋅ − = ⇒

⇒ =

3

1

1

3

3

1 1

3

1 1

3

3

32

3

2

33

3 32

23

13

3

2

2

22

2

3

3 2

2

2

3

9 2 2 2

6

7 2

6− ⇒ = − ⇒ =

−=E E

III. a) 1

9 2

9 2

9 2

9 2 9 2

9 2 9 2

9 2 9 2

7

4 4

4 4

4 4

4 4

4 4

+

⋅−

−

=− ⋅ +

− + +

∴

∴− ⋅ +

( ) ( ) ( )

( ) ( )

b) 35 3

5 35 3

3 5 35 3

5 35 3

3 5 3

4 4

4 4

4 4

4 4

4 4

−

⋅ +

+

= ⋅ +

−

⋅ +

+

∴

∴⋅ + ⋅

( )( )

( ) ( )( )

( ) ( 55 3

2

+ )

04 I. a)1

1 2

1 2 4

1 2 4

1 2 4

11 2 4

3

3 3

3 3

3 33 3

−

⋅+ +

+ +

=+ +

−

= − − −( )

( )

b) 14

25 10 4

14

5 5 2

5 2

5 2

14 5 2

7

2

3 3 3 3 3 3 2

3 3

3 3

3 3

− +

=

− ⋅

⋅+

+

=

=+

= ⋅

( ) ( )

( )

( )

( )(( )5 23 3

+

c)4 1

3 1

9 3 1

9 3 1

4 1 9 3 1

2

3

3

3 3

3 3

3 3 3−

−

⋅+ +

+ +

=− ⋅ + +( )

( )

( ) ( )

Mergulhando Fundo



LIVRO 1 | ÁLGEBRA

II. 1

4 4 3 3

4 3

4 3

4 3

2 3

2 3

2 3

4 3 2

6 2 6 6 6 2

6 6

6 6

6 6

6 6

( ) ( )

( ) (

− ⋅ +

⋅+

+

=+

+

⋅−

−

=

= + ⋅ − 33 )

III.1

729 32 3 4 256

1

3 32 3 4 21

1 32 3 2

1

1 2 2 3 2

1

1

6 4 8 8 4 8

4 4 4 4

− + −

=

− + −

=

=

− +

=

− +

=

+ 22

1 2

1 2

1 2

1 2

1 2

1 2

1 2 1 2

1

4

4

4

4 4

⋅−

−

=

=−

−

⋅+

+

=− ⋅ +

−

( )

( )

( )

( )

( ) ( )

IV. C

5

7 7 2 2

7 2

7 2

5 7 2

7 2

7 2

3 2 3 3 3 2

3 3

3 3

3 33 3

( ) ( )

( )

+ ⋅ +

⋅−

−

=

⋅ −

−

= −

d)4 1

2 1

4 2 1

4 2 14 1 4 2 1

3

3

3 3

3 3

3 3 3−

−

⋅+ +

+ +

= − ⋅ + +( )

( )( ) ( )

e)3 2

2 18 9

3 2

2 2 9 9

2 9

2 9

3 2 2

3 6 3 6 2 6 6 6 2

6 6

6 6

+

− +

=+

− ⋅ +

⋅+

+

=

=+ ⋅

( ) ( )

( )

( )

( ) ( 66 66 69

2 3

2 9+

+

= +)

( )

f)1

2 2 3

4 4 2 6 9

4 4 2 6 9

4 2 9

16 3

4 4 2 6 9

19

33

3 6 3

3 6 3

3 3

3 3 3

+

⋅− +

− +

⋅−

+

∴

∴− +

( )

( )

Resolucao 2014 9ºano Algebra l1

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Transcript of Resolucao 2014 9ºano Algebra l1