Baixe Exercícios de Radicais: Álgebra para o Ensino Médio e outras Exercícios em PDF para Matemática, somente na Docsity!
2 Álgebra
2.1. Radicais
Pág. 70
1. Seja l o comprimento do lado de um quadrado.
1.1. l = 16 = 4
l = 4 cm
1.2. l = 81 = 9
l = 9 cm
1.3. l = 0,
l = 0,1 cm
2. Seja a o comprimento da aresta de um cubo.
3 a = 8 = 2
a = 2 cm
3 a = 1000 = 10
a = 10 cm
3 a = 0,001 =0,
a = 0,1 cm
2 2 2 2 × 3 × 3 =
= 2 × 3 × 3 = 18
3.2. 2500 = 25 × 100 = = 5 × 10 = 50
4.1. 2 × 8 = 16 = 4
2 2 2 × 2 = 2 2 = 2 × 2 = 4
× = × = =
× ×
= = = × =
5.1. 12 = 4 × 3 = 4 × 3 = 2 3
5.2. 27 = 9 × 3 = 9 × 3 = 3 3
5.3. 48 = 16 × 3 = 16 × 3 = 4 3
5.4. 80 = 16 × 5 = 16 × 5 = 4 5
5.5. 600 = 100 × 6 = 100 × 6 = 10 6
2 2450 = 2 × 5 × 49 =
2 = 5 × 49 × 2 =
= 5 × 7 × 2 = 35 2
6.1. 3 2 = 9 × 2 = 18
6.2. 4 5 = 16 × 5 = 16 × 5 = 80
2 2 a 5 = a × 5 = 5 a , a > 0
2 2 2 2 4 2 3 x y z = 9 × x × y × z = 9 x y z ;
e ,
x ∈ ℝ y z ∈ℝ
= 2 + 25 × 2 − 49 × 2 + 9 × 2 =
= 2 + 25 × 2 − 49 × 2 + 9 × 2 =
= 4 × 5 − 9 × 5 + 5 − 16 × 5 =
= 4 × 5 − 9 × 5 + 5 − 16 × 5 =
= 2 16 × 3 + 3 9 × 3 − 25 × 3 − 3 =
= 2 16 × 3 + 3 9 × 3 − 25 × 3 − 3 =
= 2 × 4 × 3 + 3 × 3 × 3 × 5 × 3 × 3 =
= 2 − 3 2 + 4 × 2 =
a + a + a =
= a + a + a =
a a a
Pág. 71
3 1 1 2 1 3 : 2 2
− − × −^ −^ =
3 2
3
= × − − =
= × × − =
× ×
2 2 2
1 1
− −
− −
× ^
×
^ ^
−^ −
2 2
1
−
−
2
2
= × = =
×
7 7 7 7
7 2 7 2 7 2 2
− − − −
− − − − − − −
× ×
× × ×
2 = 12 = 144
1 2 5 3 2 1 4 7 : 7 7
− − −
× − =
2 10 1 12 7 : 7 7
− = × (^) − (^) =
10 2 12 7 × 7 : 7 =
12 12 = 7 : 7 = 1
324 2 162 2 81 3 27 3 9 3 3 3 1
2450 2 1225 5 245 5 49 49 1
2 2 x − 3 − 2 x + 1 − 2 x − 3 x + 3 =
( ) ( )
2 2 2 = x − 6 x + 9 − 2 x + 2 x + 1 − 2 x − 9 =
2 2 2 = x − 6 x + 9 − 2 x − 4 x − 2 − 2 x + 18 = 2 = − 3 x − 10 x + 25
3 2 1 1 1 3 3 3 2 2 2
−^ =^ × −^ −^ =
x x
x x x
= x − x − x + x + x − =
= x − x + x −
9.3. (^) ( )( ) ( )
2 2 2 2 2 2 2
^
x x x
( )
2 2 2 4 = 2 − x = 4 − 4 x + x =
4 2 = x − 4 x + 4
−^ −^ =^ ⇔
x x
⇔ x − = ∨ x − = ⇔
⇔ x = ∨ x = ⇔
⇔ x =
S
2 3 x − 5 x + 2 = 0 ⇔
⇔ x = ⇔
⇔ x = ⇔
⇔ x = ∨ x = ⇔
⇔ x = ∨ x =
S
11. Por exemplo:
x x
11.2. (^) ( )
x x
2 2 2 9 x − 1 = 3 x − 1 = 3 x − 1 3 x + 1
2 x − 2 − 2 x − 2 =
= ( x − 2 ) ( x − 2 − 2 )=
= ( x − 2 ) ( x − 4 )
12.3. ( ) (^) ( )
2 2 2 4 x − 5 = 2 x − 5 =
= (^) ( 2 x − (^5) )( 2 x + (^5) )
2 2 2 4 x + 12 x + 9 = 2 x + 2 × 3 × 2 x + 3 =
2 = 2 x + 3
12.5. (^) ( ) ( )( )
2 2 25 x − 10 x 3 + 3 = 5 x − 3 = 5 x − 3 5 x − 3
Pág. 72
Atividade inicial 1
1.1. (^) ( )
2 2 2 < 2 ⇔ 2 < 2
1.2. (^) ( )
3 3 2 < 2 ⇔ 2 < 2
1.3. (^) ( )
4 4 2 < 2 ⇔ 2 < 2
1.4. (^) ( )
5 5 2 < 2 ⇔ 2 < 2
2.1. ( ) (^) ( )
2 2 − 2 < − 2 ⇔ − 2 > − 2
2.2. ( ) (^) ( )
3 3 − 2 < − 2 ⇔ − 2 < − 2
2.3. ( ) (^) ( )
4 4 − 2 < − 2 ⇔ − 2 > − 2
2.4. ( ) (^) ( )
5 5 − 2 < − 2 ⇔ − 2 < − 2
Pág. 73
1.1. (^) ( ) (^) ( )
7 7 − 7 < − 2 ⇒ − 7 < − 2
1.2. (^) ( ) (^) ( )
10 10 − 7 < − 2 ⇒ − 7 > − 2
(^9 ) − 4 < 1 ⇒ − 4 < 1
(^16 ) − 4 < 1 ⇒ − 4 > 1
4 4 a < b < 0 ⇒ − a > − b
Pág. 75
2.1. (^) ( )
2 4 4 2 4 2 81 = 9 = 3 =
4 4 3 = 3 e 3 > 0
5 5 10 = 10
6 − 4 não está definido porque – 4 < 0 e 6 é par.
20 20 9 = 9 e 9 > 0
2.5. (^) ( )
7 7 14 7 2 7 7 − 2 = − 2 = − 4 = − 4
12 26 6 6 − 2 = ^ − 2 = 4
e 4 > 0
7 0 = 0
16 0 = 0
Pág. 76
4 2 4 2: 2 = 2 = 2
4 4 2 4 = 2 = 2
12 4 12:4 4:4 (^3) 3 = 3 = 3
12 9 12:3 9:3 4 3 3 = 3 = 3
8 6 8:2 6:2 4 3 2 = 2 = 2
3.6. (^) ( )
5 20 4 5 4 5 = 5 5 = 5
6 2 4 3 2 2 × 3 = 2 × 3
18 2 6 12 14 9 3 6 7 ; , , ,
a b c d = ab c d a b c d ∈ ℝ
4 2 6 = 6 = 36
3 6 3 3 6 2 27 a = 3 a = 3 a , a > 0
( )( )
a a
a a (^) a a a a
a a a a a a a
( )( )
( )( )
− − +^ −
( )
− + ^ − − + −
( ) (^ )
2
− −^ −^ −^ +
3 2 1 (^3 2 1 )^2
×
( )
( )( )
( ) ( )
(^2) ( 2 1 2 2 ) 2 2 2 2 2
Pág. 84
3
( )
( )
2 3 3 2
3 3 3
+ × +
3 3 81 2 9 4
3 3 3 = 3 × 3 + 2 9 + 4 =
3 3 = 3 3 + 2 9 + 4
3 3
( ) ( ) ( )
( ) ( )
2 2 3 3 3 3
3 3 3 3
+ × − + −
( )
3 3 3 2 9 6 4
3 3 3 2 9 2 6 2 4
3
3
( ) (^ )^ (^ )
( ) (^ )
(^2 ) (^3 )
(^3 ) 3
+ × − + −
( ) (^) ( )
(^3 2 3 ) 5 24 24 1 3 3 3 8 24 1
− + × − +
( )
2 3 3 3 9 2 24 1
× − +
3 3 4 9 24 1
4
4
( ) ( ) ( )
( )
3 2 4 4 4 4 2 3
4 4 4
+ + − + × +
( )( )
4 4 3 4 2 4 2 1 2 2 2 1
4 4 4 3 4 2 4 4 3 4 2 4 = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 1 =
4 3 4 = 2 + 2 3 + 2 2 + 2 2 + 1 =
4 4 = 3 + 2 2 + 2 8 + 2 2
(^4 44 )
4 4
( ) ( ) ( ) ( ) ( )
( ) ( )
3 2 2 3 4 4 4 4 4 4
4 4 4 4
+ − + × − + −
( ) ( )
2 3 4 3 4 2 4 2 4 4 4 2 6 3 3 3 3 3 3
−^ +^ −
4 4 4 = 3 − 27 + 3 9 − 3 3
Pág. 85
14.1. (^) ( )
4 3 3 3 4
3 6
2 (^3 ) 6
× ×
= × − =
3 6
×
6 3 3 6
3 3 6 2 2 2 4 2
3 6 2 = 2 2 − 4 2 =
3 = − 2 2
3 3 2 2 A − B = A − B A + AB + B
3 A = 9 e B = 2
3 3 2 2 A − B = A − B A + AB + B
3 3 A = 3 e B = − 2
3 3 2 2 A − B = A − B A + AB + B
3 A = 24 e B = − 1
4 4 3 2 2 3 A − B = A − B A + A B + AB + B
4 A = 2 e B = 1
4 4 3 2 2 3 A − B = A − B A + A B + AB + B
4 4 A = 3 e B = − 9
( )
3
3 4
×
2 4 2
4 3
× × × ×
8 3
4 3
× ×
8 4 8 3
8 6
× ×
4 3 8 8 6
×
3 3 3
(^3 5 )
− ×
× × ×
3 3 3 33
4 6 3
× −
× × ×
3 3 3
6 3 6 3 6 3
× × ×
3 3
6 3 3 2 6 6 6 2
× × ×
3
3
× ×
4
8 2 3
× =
4 4 2 6 3
(^4 6 )
+ ×
= × =
4 2 3 6 4 2
= × =
6 4
= × =
2 6 6 4 4
×
= × = × =
4 3 3 6 4 6 4
= × = × =
4 2 6 6 = 3 × 3 = 3 × 3 =
6 3 6 6 4 3 2 = 3 × 3 = 3 = 3
Pág. 86
15.1. 3 + 2 2 = 3 + 2 × 1 2 =
( )
2 2 = 2 + 2 2 + 1 =
( )
2 = 2 + 1
( )
2 3 + 2 2 = 1 + 2 = 2 + 1 = 1 + 2
15.2. 3 − 2 2 = 3 − 2 × 1 2 = ( )
2 2 1 + 2 = 3
( )
2 2 = 2 − 2 × 1 × 2 + 1 =
( )
2 = 2 − 1
( )
2 3 − 2 2 = 2 − 1 = 2 − 1
15.3. 37 − 20 3 = 37 − 2 × 10 3 =
( )
2 2 = 5 − 2 × 10 3 + 2 3 =
( )
2 = 5 − 2 3
( )
2 37 − 20 3 = 5 − 2 3 = 5 − 2 3
15.4. 21 − 8 5 = 21 − 2 × 4 5 =
( )
2 2 = 4 − 2 × 4 5 + 5 =
( )
2 = 4 − 5
( )
2 21 − 8 5 = 4 − 5 = 4 − 5
− = − × =
( )
2 (^2 1 ) 2 2 2 2 2
= − × + =
2 1 2 2
15.6. 5 + 2 6 = 5 + 2 × × 1 6 =
( ) ( )
2 2 = 3 + 2 × 1 × 6 + 2 =
( )
2 = 3 + 2
( )
2 5 + 2 6 = 3 + 2 = 3 + 2
Pág. 87
16.1. 51 + 14 2 = 51 + 2 × 7 2 =
( )
2 2 = 7 + 2 × 7 2 + 2 =
( )
2 = 7 + 2
( )
2 51 + 14 2 = 7 + 2 = 7 + 2
16.2. 22 − 8 6 = 22 − 2 × 4 6 =
( )
2 2 = 4 − 2 × 4 6 + 6 =
( )
2 = 4 − 6
( )
2 22 − 8 6 = 4 − 6 = 4 − 6
16.3. 33 + 20 2 = 33 + 2 × 10 2 =
( )
2 2 = 5 + 2 × 10 2 + 2 2 =
( )
2 = 5 + 2 2
( )
2 33 + 20 2 = 5 + 2 2 = 5 + 2 2
9 + 32 = 9 + 16 × 2 = 9 + 4 2 =
108 2
54 2
27 3
9 3
3 3
1 3
144 2
72 2
36 2
18 2
9 3
3 3
1
2 2 1 + 2 = 3
3 − 2 2 > 0
2 − 1 > 0
ab = 10 3
2 2 3 10 + 3 = 10
2 3 5 + 2 3 = 25 + 12 = 37
ab = 4 5
2 2 4 + 5 = 16 + 5 = 21
1 2 2
ab =
2 1 2 1 9 2 2 2 4 4
+^ =^ +^ =
ab = 1 × 6 = 2 × 3
2 2 6 + 1 = 7
2 2 3 + 2 = 5
ab c = 7 2
2 2 7 + 2 = 51
ab c = 4 6
2 2 4 + 6 = 22
ab c = 10 2
= 5 × 2 2
2 2 2 10 + 2 = 10
2 2 5 + 2 2 = 33
Pág. 89
20. Área da coroa circular: A C
2 2 = π − π C A R r
Pelo Teorema de Pitágoras: 2 2 2 2 2 r + r = R ⇔ R = 2 r
Então: 2 2 2 = π × 2 − π = π C A r r r
Como AC = π a , vem:
2 2 π r = π a ⇔ r = a ⇔ r = a
Como o lado do quadrado é igual a 2 r , o perímetro é:
P = 4 × 2 r = 8 a
Logo, P = 8 a cm.
21. V esfera = (^) ( )
π ×
= 4 π 3
V semiesfera = 2 π 3
Volume do cilindro de
altura 3 :
V clinidro = (^) ( )
2 = π 3 × 3 = 3 π 3
V água = 3 π 3 − 2 π 3 = π 3
π h = π ⇔ h =
h = cm
Pág. 91
Atividades complementares
3 3 1 1 1 1
2 3 2 3
4 4 1 1 1 1
4 2 4 2
2 3 3 2 2 2 3 3
2 1 1 2 1 1 3 3
23.1. (^) ( ) ( )
5 5 − 5 < − 3 ⇔ − 5 < − 3
6 6 2 3 2 3
2 3 2 3
− − ^ ^ ^ −
8 8 1 1 1 1
3 4 3 4
7 7 7 8 7 8
8 7 8 7
(^3 ) x = 8 ⇔ x = 8 ⇔ x = 2
S = { 2 }
(^3 ) x = − 27 ⇔ x = − 27 ⇔ x = − 3
S = { − 3 }
4 x = − 3 é impossível
S = { }
4 4 4 x = 3 ⇔ x = − 3 ∨ x = 3
{ }
4 4 S = − 3 , 3
4 4 4 4 4 4 4 x = 10 000 ⇔ x = 10 ⇔ x = 10 ∨ x = 10 ⇔
⇔ x = − 10 ∨ x = 10
S = { −10 , 10}
6 x = − 4 , é impossível
S = { }
7 7 x = − 1 ⇔ x = − 1 ⇔ x = − 1
S = { − 1 }
5 5
x = − 32 ⇔ x = − 32 ⇔ x = − 2 ( )
5 2 = 32
S = { − 2 }
8 x = 0 ⇔ x = 0
S = { 0 }
x = − ⇔ x = ⇔ x = − ( )
3 4 = 64
S
25.1.^3
4 4 7 = 7
8 8 − 1 = 1 = 1
7 − 1 = − 1
7 7 2 = 2
25.6. (^) ( )
6 6 12 6 2 2 = 2 = 4
(^8 8 ) 8 − 8 = 8 = 8
7 7 − 5 = − 5
25.9.^5
5 2 = 32
15 0 = 0
4 4 4 0,0001 = 0,1 =0,
3 3 3 0,000000008 = 0,002 =0,
2 2 2 4 3 3 9
× = =
6 2 6:2 2:2 3 7 = 7 = 7
(^2 4 ) 2 2 3 3 9 , 0 × a = a = a a >
26.4. (^) ( )
2 (^5 2) 2 5× 2 10 4 2 a b = a b = a b
15 5 10 15:5 1 2 3 2 a b = a b = ab
3 6 3 9 ab = a b , a > 0 e b > 0
27. Por exemplo:
4 2 6 3 8 4 2 = 2 = 2 = 2
3 6 2 9 3 12 4 5 = 5 = 5 = 5
9 3 3 6 2 12 4 3 = 3 = 3 = 3
(^3 2 6 2 4 9 3 6 124 ) 2 x = 2 x = 2 x = 2 x , x > 0
4 2 8 2 4 12 3 6 16 4 8 2 x = 2 x = 2 x = 2 x , x > 0
2 2 4 3 6 4 8 5 10 15 20 3 6 9 12 = = = ; > 0, > 0 e > 0
xy x y x y x y x y z z z z z
28.1. m.m.c. (3 , 2) = 6
3 6 2 6 5 = 5 = 25
6 3 6 2 = 2 = 8
28.2. m.m.c. (4 , 8) = 8
4 8 8 12 = 144 = 144
8 8 144 = 5
28.3. m.m.c. (3 , 4) = 12
3 12 4 12 7 = 7 = 2401
4 12 3 12 2 = 2 = 8
28.4. m.m.c. (2 , 4) = 4
( )
2 2 4 2 4 4 2 a = 2 a = 4 a , a > 0
4
b b >
pág. 92
29.1. m.m.c. (3 , 5 , 15) = 15
(^3 15 ) 2 = 2 = 32
(^5 3 159 ) 2 = 2 = 512
(^15 2 ) 7 = 49
15 15 15 32 < 49 < 512
Logo,
3 15 2 5 3 2 < 7 < 2.
29.2. m.m.c. (3 , 6 , 9) = 18
3 18 6 18 2 = 2 = 64
6 2 18 6 18 3 = 3 = 729
9 2 18 4 18 5 = 5 = 625
18 18 18 64 < 625 < 729
Logo,
3 9 2 6 2 2 < 5 < 3.
29.3. m.m.c (4 , 6 , 8) = 24
4 24 6 24 2 = 2 = 64
6 2 24 8 24 2 = 2 = 256
8 24 3 24 5 = 5 = 125
24 24 24 64 < 125 < 256
Logo,
4 8 6 2 2 < 5 < 2.
29.4. m.m.c. (2 , 3 , 4) = 12
12 6 12 3 = 3 = 729
3 2 12 8 12 3 = 3 = 6561
4 2 12 6 12 2 = 2 = 64
12 12 12 64 < 729 < 6561
Logo,
4 2 3 2 2 < 3 < 3.
30.1. 2 × 32 = 64 = 8
30.2. 3 × 4 = 2 3
30.3. 2 × 8 = 16 = 4
3 3 3 2 × 5 = 10
4 4 4 2 × 8 = 16 = 2
30.6. (^) ( )
6 2 3 3 6 5 2 × 32 = 2 × 2 =
(^6 3 10 ) = 2 × 2 = 2 × 2 =
(^6 12 6 ) = 2 × 2 = 4 2
30.7. (^) ( )
3 3 4 12 4 12 3 2 × 8 = 2 × 2 =
12 4 9 12 12 12 = 2 × 2 = 2 × 2 = 2 2
4 8 6 4 4 3 x × x = x × x =
4 3 4 4 = x × x = x = x , x > 0
3 6 2 6 3 6 2 3 2 5 × 3 2 = 6 × 5 × 2 = 6 5 × 2 =
6 6 = 6 25 × 8 = 6 200
4 5 4 4 10 4 10 x × x = x × x = x × x =
4 11 4 8 3 2 4 3 = x = x × x = x x , x > 0
6 5 4 3 12 10 12 9 12 19 2 × 2 = 2 × 2 = 2 =
12 12 7 12 = 2 × 2 = 2 128
× = × × =
= × × × =
= × × × =
(^15 6 15 5 ) = 2 × 2 = 2 =
15 = 2048
31.1. (^) ( )( ) ( )
2 2 2 − 3 2 + 3 = 2 − 3 = 4 − 3 = 1
31.2. (^) ( )
2 2 + 3 = 2 + 2 2 3 + 3 = 5 + 2 6
31.3. (^) ( )
2 2 2 − 5 3 = 8 − 2 × 2 2 × 5 3 + 75 =
31.4. (^) ( 2 3 − (^2) )( 3 − (^2) )=
= 2 × 3 − 2 6 − 6 + 2 =
31.5. (^) ( ) ( ) 4 4 4 4 4 8 2 − 2 = 8 2 − 4 =
4 4 = 16 − 32 =
4 4 4 5 = 2 − 2 =
4 4 = 2 − 2 × 2 =
4 = 2 − 2 2
31.6. (^) ( )
2 3 3 2 81 − 3 3 =
( )
2 (^3 4 ) = 2 3 − 3 3 =
( )
2 3 3 = 2 × 3 × 3 − 3 3 =
( ) ( )
2 2 3 3 = 3 3 = 9 3 =
3 = 9 9
= 9 × 7 − 5 4 × 7 + 16 × 7 − 4 36 × 7 + 2 64 × 7 =
= 3 7 − 5 × 2 7 − 4 7 − 7 × 6 7 + 2 × 8 7 =
= × + × − × × × =
= × + × − × − × =
3 2 3 2 3 2 3 2
(^3 33 2 3 2 )
×
× ×
3 3 2 3 2 3 2 3 (^3 33 2 )
×
×
4 4 3 4 3 4 3 4 (^4 44 3 )
×
×
6 5 6 5 6 5 6 5
(^6 66 5 )
×
×
7 3 7 3 7 3 7 3
(^7 7 4 7 3 )
× × ×
× ×
( )
( )( )
6 6 2^7 6 2^ (^7 )
− − +^ −
( ) ( )
( )( )
− − +^ −
( )( )
( )( )
( )
( )( )
30 30 3^2 3 30 3^ (^2 3 )
− − +^ −
( ) ( )
( )
( )( )
a a b a
a b (^) a b a b
, 0 e 0 4
a ab a b a b
( )
( )( )
a a a a
a a (^) a a a a
( )
(^2) ( 2 ) , 0 2 2
a a a a a a a a a
3
( )
( )
2 3 3
3 3 3
3 3 4 2 2 4
3 3 4 2 2 4
3 3
( ) ( ) ( )
( ) ( )
2 2 3 3 3 3
3 2 3 3
+ × − + −
( )
(^3 2 ) 2 5 10 4
( )
4
( ) ( ) (^ )^ ( ) (^ )^ (^ )
( ) (^ )
(^3 2 2 ) 4 4 4
(^4 )
+ × − + × − + −
( )
(^4 3 42 ) 6 2 2 2 1
4 4 = 6 8 + 6 2 − 6 2 − 6
4
4 4
( ) ( ) ( )( ) ( )
( ) ( )
3 2 2 3 4 4 4 4 4 4
4 4 4 4
4 3 4 2 4 4 4 2 4 2 4 4 3 4 3 3
4 4 4 4 = 64 + 48 + 36 + 27
41.1. (^) ( )( ) ( )
2 3 − 5 3 + 5 − 2 − 3 =
= 9 − 5 − (^) ( 4 − 4 3 + (^3) )=
41.2. (^) ( ) (^) ( )
2 2 2 − 2 − 6 − 4 2 =
= 4 − 4 2 + 2 − (^) ( 6 − (^4 2) )=
41.3. (^) ( )( ) ( )
2 3 2 − 2 3 3 2 + 2 3 + 1 + 2 2 =
( ) ( )
2 2 = 3 2 − 2 3 + 1 + 4 2 + 8 =
41.4. (^) ( )
2 2 1 2 1 2 2 1 2 1 2 2
^ ^
−^ −^ −^ ^ +^ =
( )
= − + × − − + + =
( )
( )
= − − + − − ×
3 3 2 2 A − B = A − B A + AB + B
(^3 ) A = 2 e B = 2 = 8
3 3 2 2 A − B = A − B A + AB + B
3 3 A = 5 e B = − 2
4 4 3 2 2 3 A − B = A − B A + A B + AB + B
4 A = 2 e B = − 1
4 4 3 2 2 3 A − B = A − B A + A B + AB + B
4 4 A = 4 e B = 3
41.5. (^) ( )( )
2 3 2 4 1 1 3 2 3 3 2
− −^ −^ −^ +^ =
( )( )
= 4 + 4 3 + 3 − (^) ( 1 − (^3) )( 31 + (^12 3) )=
= 7 + 4 3 − (^) ( 31 + 12 3 − 31 3 − (^36) )=
( )
2 2 1 3 2 2 3 2 3
( )
( )
3 2 6 4 4 x − a : 2 x − a =
x a x a x a
3 3 3 128 − 16 + 54 =
3 3 3 = 64 × 2 − 8 × 2 + 27 × 2 =
3 3 3 3 3 3 = 4 × 2 − 2 × 2 − 3 × 2 =
3 3 3 = 4 2 − 2 2 + 3 2 =
3 = 5 2
3 4 3 4 3 4 16 x + 54 x − − 128 x =
3 3 3 3 3 3 = 8 x × 2 x + 27 x × 2 x + 64 x × 2 x =
(^3 3 3 3 ) 3 3 3 = 2 x × 2 x + 3 x × 2 x + 4 x × 2 x =
3 3 3 = 2 x × 2 x + 3 x × 2 x + 4 x × 2 x =
3 = 2 x + 3 x + 4 x 2 x =
3 = 9 x 2 x
3 3 3
6 6
× ×
( )
2 (^6 )
6 6
×
2 2 6 24 6 24 3 8 2 9 3
(^4 3 1 433 )
(^3 3 )
− − × = =
×
( )
(^4 3 2 ) 4 2 3 10 4 3 5
= = × =
4 4 3 12 6 10
− = = =
1 2 1 2 2 2 2 2
− − = = × =
2 4 + 2 3 = a + b
( )
2 2 4 + 2 3 = 3 + 2 3 + 1 =
( )
2 = 3 + 1
( )
2 4 + 2 3 = 3 + 1 = 3 + 1
2 36 − 16 2 = a − b
( )
2 2 36 − 16 2 = 4 2 − 16 2 + 2
( )
2 = 4 2 − 2
( )
2 36 − 16 2 = 4 2 − 2 = 4 2 − 2 > 0
2 7 + 2 10= a + b
( ) ( )
2 2 7 + 2 10 = 5 + 2 10 + 2 =
( )
2 = 5 + 2
( )
2 7 + 2 10 = 5 + 2 = 5 + 2
2 10 − 4 6 = a − b
( )
2 2 10 − 4 6 = 6 − 4 6 + 2 =
( )
2 = 6 − 2
( )
2 10 − 4 6 = 6 − 2 = 6 − 2 = 6 − 2 > 0
2 29 + 12 5 = a + b
( )
2 2 29 + 12 5 = 3 + 12 5 + 2 5 =
( )
2 = 3 + 2 5
( )
2 29 + 12 5 = 3 + 2 5 = 3 + 2 5 = 2 5 + 3
2 44 + 16 7 = a + b
( )
2 2 44 + 16 7 = 4 + 16 7 + 2 7 =
( )
2 = 4 + 2 7
( )
2 44 + 16 7 = 4 + 2 7 = 4 + 2 7 = 2 7 + 4
2 6 − 2 5 = a − b ab = 5 = 1 × 5
( ) ( )
2 2
2 2 5 + 1 = 6
( )
2 = 5 − 1
( )
2
( )( )
( )( )
−^ −^ −^ +
ab = 3
2 2 3 + 1 = 4
ab = 8 2 = 2 × 4 2
2 2 8 + 2 = 66
(^2 ) 4 2 + 2 = 36
ab = 10 = 5 2
2 2 5 + 2 = 7
ab = 2 6
(^2 ) 6 + 2 = 10
ab = 6 5 = 3 × 2 5 =...
2 2 6 + 5 = 41 ≠ 29
2 2 3 + 2 5 = 29
ab = 8 7 = 4 × 2 7 =...
2 2 8 + 7 = 71
2 2 4 + 2 7 = 44
Altura da pirâmide:
2 2 2 2
2
h a a
⇔ h = a − a ⇔
⇔ h = a − a ⇔
⇔ h = a ⇔ h = a ⇔
⇔ h = a ⇔ h = a
V octaedro = 2 × V pirâmide
base
2 altura 3
= × × A × =
= × a × a =
= a
48. = tg 30°
CE
EB
= EB × ⇔
⇔ 6 2 = EB × 3 ⇔
⇔ EB = ⇔
EB
EB
EB
2 2 2 BC = EB + CE
( ) ( )
2 2 2
2
BC
BC
BC
⇔ BC = 16 × 2 ⇔ BC = 4 2
Perímetro:
AD DC CB AB
Logo, o perímetro do trapézio é (^) ( 10 2 + (^2 6) )cm.
3 3 a = 2 50 ⇔ a = 2 50
3 ⇔ a = 4 × 50 ⇔
6 ⇔ a = 200
O comprimento da aresta do cubo é
6 200 cm.
49.2. V pirâmide (^) cubo
= × V =
= × =
O volume da pirâmide [ ABCDH ] é
cm 3 .
2 2 2 HC = HG + GC
( ) ( )
2 2 2 6 6 HC = 200 + 200 ⇔
( )
2 3 6 ⇔ HC = 2 × 200 ⇔
6 2 ⇔ HC = 2 200 ⇔
3 ⇔ HC = 2 × 200 ⇔
3 3 ⇔ HC = 2 × 200 ⇔
6 3 ⇔ HC = 2 × 2 × 100 ⇔
6 4 2 ⇔ HC = 2 × 10 ⇔
3 2 ⇔ HC = 2 × 10 ⇔
3 ⇔ HC = 40 cm
2 2 2 HB = DB + HD
( ) ( )
2 2 2 3 6 HB = 40 + 200 ⇔ DB = HC
2 3 2 6 2 ⇔ HB = 40 + 200 ⇔ 2 3 3 ⇔ HB = 1600 + 200 ⇔ 2 3 3 ⇔ HB = 8 × 200 + 200 ⇔ 2 3 3 ⇔ HB = 2 200 + 200 ⇔
(^2 ) ⇔ HB = 3 200 ⇔
3 ⇔ HB = 3 200 ⇔
3 ⇔ HB = 9 × 200 ⇔
6 ⇔ HB = 1800
6 BH = 1800 cm
Pág. 95
50. = cos30°
OE
EM
= EM × ⇔
⇔ EM = ⇔
⇔ EM =
tg 4 3 3
OM OM
OM
OE
AB = OM =
A base
2 (^2 8 3 64 3 )
×
AB
A face
×
×
AM EM
A total
= + × = = =
A área total da superfície da pirâmide é 64 cm 2 .
51. BC = a
2 2 2 BG = a + a ⇔
2 BG = 2 a ⇔ BG = 2 a ⇔
[ ]
= 3 × 2 = 3 2
BGE P a a
51.2. O triângulo [ BGE ] é equilátero de lado 2 a.
( )
2 2 2 2 2 2
h a a
⇔ h = a − a ⇔
⇔ h = a ⇔
⇔ h = a ⇔
⇔ h = a ⇔ h = a
[ ]
2
×
BGE
a a
A a
A área do triângulo [ BGE ] é
a.
51.3. V pirâmide [ ]
= A EBF × FG
3 1
×
= × × =
a a a a (1)
Seja h a altura pedida:
V pirâmide [ ]
= A (^) BGE × h
2 1 3 2 3
= × =
a h a h (2)
Como (2) = (1), vem: 2 3 3 2 3 2
a h a a a h a h a
a a h h h a
A altura da pirâmide [ EBGF ] é
a.
52. Sejam A, B e C os centros das circunferências interiores e O o
centro da circunferência exterior.
Seja M o ponto médio de [ AB ] e r o raio.
AC = 2 r e AM = r
Como o triângulo [ ABC ] é equilátero: 2 2 2 AC = AM + MC
(^2 2 2 22 2 ) 2 r = r + MC ⇔ MC = 4 r − r ⇔ MC = 3 r ⇔
⇔ MC = 3 r
O ponto O também é o centro da circunferência circunscrita
ao triângulo [ ABC ]. Como o triângulo é equilátero, então este
ponto também é o ponto de interseção das medianas. Logo:
2 2 3 3 3
OC = MC = r
Seja [ ON ] o raio da circunferência exterior que passa em C.
Então, OC + CN é igual ao raio da circunferência.
2 3 6 2 3 3 18 3
r + r = ⇔ r + r =
( )
r r
( )
( )( )
18 2 3 3 18 2 ( 3 3 )
+ −^ −
r r
⇔ r = 6 2 ( 3 − (^3) ) ⇔ r = 12 3 − 18
O raio de cada uma das circunferências é (^) ( 18 − (^2 3) )cm.
53. A altura da caixa deve ser igual ao diâmetro dos sabonetes, ou
seja, 6 cm.
Para determina as dimensões da base da caixa vamos
considerar o seguinte esquema onde se visualiza a planta da
caixa com os sabonetes. 2 2 2 PR = QR + PQ Teorema de
Pitágoras 2 2 2 6 = 3 + PQ ⇔
⇔ PQ = 36 − 9 ⇔
⇔ PQ = 3 3
Os triângulos [ PSC ] e [ RQP ]
são semelhantes por terem dois ângulos iguais
( )
CSP = RQP = 90 ° e QPR = PCS
CS QR
PQ PS
, ou seja,
CS
CS
Podemos finalmente, concluir que o lado do triângulo
equilátero [ ABC ] é igual a:
CS + ST + TB = 3 3 + 6 + 3 3 =
A caixa dos sabonetes tem a
forma de um prisma
triangular regular com 6 cm
de altura e (^) ( 6 + (^6 3) )cm de
aresta da base.
Pág. 96
Avaliação 1
( ) ( )
(^3 ) 4 2 4 3 4 2 4 3 2 2 2 2
( ) ( )
3 3 4 4 4 4 4 8 4 8
( )
4 4 4 3 4 3 ⇔ 4 > 8 ⇔ − 4 > − 8 ⇔
( )
4 4 4 3 4 3 ⇔ 4 > 8 ⇔ 4 < 8 ⇔
⇔ ( F ⇔ V) ⇔F
Resposta: (C)
12.1. (^) ( ) (^) ( )
2 2 2 x − 2 + x = 5 ⇔
2 2 ⇔ x − 4 x + 4 + x − 5 = 0 ⇔ 2 ⇔ 2 x − 4 x − 1 = 0 ⇔
⇔ x = ⇔ x = ⇔
⇔ x = ⇔ x = ⇔
⇔ x =
Como x > 0, temos
x =.
12.2. A quadrado ( )
2 2 2 2 6 2 6 4 2 2 2 2
x
2 6 2 6 4 6 4 10 4 6 5 6 2 4 4 2
A área do quadrado é
cm 2 .
13. Seja a a aresta do cubo.
2 2 2 EB = a + a
2 2 2 EC = EB + BC 2 2 2 2 2 EC = a + a + a ⇔ EC = 3 a ⇔
⇔ EC = 3 a (^) ( EC > 0 e a > (^0) )
Como EC = 12 , temos:
a = ⇔ a = ⇔ a = ⇔ a =
3 V cubo = 2 = 8
V pirâmide
= × × = , logo
V = − =
O volume da parte do cubo não ocupada pela pirâmide é
16
cm 3 .
6 3 2 x − 3 x − 3 = 0
Para
6 x = 3 :
( ) ( )
6 3 6 6 2 3 − 3 × 3 − 3 = 0 ⇔
6 3 ⇔ 2 × 3 − 3 × 3 − 3 = 0 ⇔
3 ⇔ 6 − 3 × 3 − 3 = 0 ⇔
Para 6
x = − :
6 3
6 6
× − − × − − = ⇔
3 6
⇔ × + × − = ⇔
⇔ + × − = ⇔
⇔ + × − = ⇔
2 7 − 4 3 = a − b
( )
2 2 7 − 4 3 = 2 − 4 3 + 3 =
( )
2 = 2 − 3
( )
2
De igual modo:
( )
2 7 + 4 3 = 2 + 3 = 2 + 3
16. AB = 6 cm
AC = BC = 9 cm
Seja M o ponto médio de [ AB ]. 2 2 2 AC = AM + MC 2 2 2 9 = 3 + MC ⇔
⇔ MC = 81 − 27 ⇔
⇔ MC = 54 ⇔ ( MC > 0 )
⇔ MC = 3 6
MO = MC − r ⇔
⇔ MO = 3 6 − r 2 2 2 AM + MO = r
( )
2 2 2 2 9 + 3 6 − r = r ⇔ 9 + 54 − 6 6 r + r = r ⇔
⇔ r = ⇔ r = ⇔ r = ⇔
× ×
r r r
O raio da circunferência circunscrita ao triângulo é
cm.
17. Como AE = DC = 3 e ED // AC , então [ ACDE ] é um trapézio
isósceles.
Logo,
EDC = AED = 120 °
Seja M ∈[ AD ] tal que DM ⊥ AC e
MDC = 30 °
Seja h a altura do trapézio.
cos30 3 2 3
= ° ⇔ = × ⇔ =
DM
h h DC
sin 30 3 2 2
= ° ⇔ = × ⇔ =
MC
x x DC
3 2 6 2 9 2
AC = ED + x = + × =
Se
BAC = 30 ° e ACB = 60 °, então
CBA = 90 °, ou seja, o
triângulo [ ABC ] é retângulo em B.
ab = 2 3
2 2 2 + 3 = 7
sin 30 9 2 2
BC BC
BC
AC
cos 30 9 2 2
= ° ⇔ = × ⇔ =
AB
AB AB
AC
A = A trapézio + A triângulo =
×
= × + =
A área do pentágono [ ABCDE ] é
cm 2 .
18.1. Sendo 2 r é a diagonal espacial do cubo:
2 2 2 d = a + a ⇔
(^2 2 ) ⇔ 2 r = d + a ⇔
(^2 2 2 ) 2 r = a + a + a ⇔
2 2 ⇔ 4 r = 3 a ⇔
⇔ r = a ⇔
⇔ r = a ⇔
⇔ r = a , a > 0
π 3
V =
Como
r = a :
3 2 (^4 3 4 3 ) π π 3 2 3 2 2
= × ⇔ = × × × ⇔
V a V a
π 3 4 2
⇔ V = × × a ⇔
π 2
⇔ V = × a ⇔
3 ⇔ 2 V = π 3 a ⇔
3
3
π 3
π 3 3
×
V
a
V
a
3
3 π
V
a
2.2. Potências de expoente racional
( )
1 2 0
3 2 4 3 4 3 4
− −^ − +
× × −^ ×
2 2 2 AB + AE = BE AB = AE
( )
4 2 2 2 2 2 4 2 2 2 2 2 2
AB = ⇔ AB = ⇔ AB =
2 AB é a área do quadrado [ ABDE ].
Área de [ ABCE ]
( )
1 (^3 ) 3
− = = = × =
3 2 3 2
− = ×
A área do quadrilátero é
3 2 3 2
×
cm 2 .
2 2 2 AC = AB + BC
( )
(^2 ) 4 2 32 = 2 AB ⇔
4 2 (^2 4 )
⇔ AB = ⇔
2 ⇔ AB = 2 32 ⇔
⇔ AB = 2 32 ⇔
⇔ AB = 4 × 32 ⇔
4 2 5 4 4 3 ⇔ AB = 2 × 2 ⇔ AB = 2 × 2 ⇔
4 ⇔ AB = 2 8
Altura do triângulo, h :
( ) ( )
2 2 2 4 4 2 4 2 4 2 h + 8 = 2 8 ⇔ h = 4 8 − 8 ⇔
2 4 ⇔ h = 3 8 ⇔ h = 9 × 8 ⇔ h = 72
[ ]
4 4 2 8 72
AED
A
×
4 4 3 = 8 × 72 = 2 × 2 × 36 =
4 4 2 2 = 2 × 6 = 2 × 6 =
1 2 = 2 6 = 2 × 6
A área do triângulo é
1 2 2 6
×
cm 2 .
Pág. 105
13.1. Altura do prisma:
h = m
A base = 1,92 m 2
Aresta da base: a
a = = =
6 192 2 3
×
3 2 3 8 3
×
A lateral
= × a × h = × ×
×
A área da superfície lateral do prisma é
m 2 .
13.2. Seja r o raio da esfera.
r = a ⇔ r = ⇔ r =
×
h
r
14. a = AF = 24
b = CD = 9
( )
3 (^2 2 ) 3 3 C 24 9
( )
3 2 2 3 3 2 24 9
+^ =
( )
3 2 2 3 3 2 24 3
( )
3 3 3 3 3 2 = 2 × 3 + 3 × 3 =
( )
3 3 3 2 = 2 3 + (^3 3) ( ) ( )
1 3 (^3 ) 3 2 3 3 5 3 5 3
= = × =
( ) ( )
1 1 (^3 2 ) = 5 × 3 = 5 × 5 × (^3) ( )
1 1 (^2 ) = 5 × 15 =
1 2 = 5 × 15
Pág. 107
Atividades complementares
15.1. (^) ( )
1 1 3 3 3 1 8 = 2 = 2 = 2
0, 0,5 2 1 1 1 1 1
=^ ^ =^ =
1 1 1 3 3 3 3 3
3
× ^ =^ =^ = (^)
15.4. (^) ( )
0, 0,25 4 1 10 000 = 10 = 10 = 10
1 2 1, 44 = 1, 44 =1, 2
0, 0,2 0,2 5 5 0,2 1
5
× =^ =^ ^ =^ =^ = (^) (^)
15.7. (^) ( )
1 1 6 66 64 = 2 = 2
0, 0,25 0,25 (^4) 4 0,
4
^ ×
=^ =^ ^ =^ =
1 2 5 = 5
3 3 1 2 1 1
=^ =
1 1 2 2 1 2
x , x x x x
− + = = (^) = ∈
1 3 2 3 3 2 2 4 6 2 2 2
× = = × =
192 2 96 2 48 2 24 2 12 2 6 2 3 3 1
2.2. Potências de expoente racional
1 2
2
5 5 1 2 2 2 2
2
a a a a a a a
−
= = = = ∈ ℚ
1 (^3 ) 7 = 7 17.2.
1 (^4 ) 20 = 20
1 5 5
1 8 8
, a 0 a a
0,
1 20 ab = ab^20 , ab > 0
18.1. (^) ( )
2 2 3 3 3 2 1000 = 10 = 10 = 100
18.2. (^) ( )
0, 0,4 5 2 32 = 2 = 2 = 4
18.3. (^) ( )
4 4 4 3 3 3 3 3 4 8 2 2 2 16
× = = = =
18.4. (^) ( )
2 2 2 6 3 6 3 3 4 64 2 2 2 16
× = = = =
18.5. (^) ( )
1, 1,5 2 2 1,5 3 9 3 3 3 27
× = = = =
18.6. (^) ( )
5 5 4 4 4 5 16 = 2 = 2 = 32
3 4 4 3 4 2 = 2 = 8
2 3 3 2 3 4 = 4 = 16
2 (^3 2 ) 2 = 2 20.2.
5 (^4 5 ) 3 = 3
20 5 20 5 4 a = a = a
1 (^3 3 6 ) 2 2 = 8 = 8 = 2 = 2 = 2
1 3 15 5 1 3 53 1 15 1 155 5 5 5 5 2 2 8 8 18
= × = × = =
1 (^2 2 2 ) 3 3 3 3
× × = = =
1 2 5 = 5 21.2.
3 (^3 ) 5 = 5
4 (^3 3 3 ) 625 = 25 × 25 = 5 = 5
3 (^3 4 3 ) 5 = 5 = 5 21.5.
5 (^3 4 5 12 5 ) 5 = 5 = 5
1 3 3 2 2 (^12) 5 5 5
× × = =
22.1. (^) ( )
1 1 2 3 3 2 3 3 2
− − = (^) = =
1 1 2 2 1 2
^ −
22.3. (^) ( )
1 5 3 5 3 3 5
− − = =
3 4 4 3 3 4
− = =
2 1 2 1 7 3 2 3 2 6 5 5 5 5
× = =
3 1 3 1 11 5 2 5 2 10 10 :10 10 10
− − − = =
1 3 13 1 3 13 13 13 3 4 12 3 4 12 12 12 2 2 : 2 2 : 2 2 : 2 1
× = = =
1 1 1 1 5 3 2 3 2 6 5 5 6 6
5 5 5 5 6 6 6 6
× ^ ^ ^ = = = (^) =
Pág. 108
1 1 1 2 2 2 3 9 × 3 = 27 = 3 = 3 3
1 1 2 2 1 2 1 5 5 5 5 2 5 20 : 4 20 : 4 5 5 5 25
=^ =^ =^ =
^
2 4 2 4 3 3 2 3 3 2 2 2 2 6 6 : 3 6 : 3 6 : 3 6 : 3
− − − − (^) − − − − − × = = = =
− = =
1 1 2 1 7 3 6 6 9 : 3 : 3
− − − =
1 1 7 6 6 6 9 : 3 : 3
− − − = =
1 7 6 6 9 : 3 : 3
− − = =
1 7 1 7 6 6 6 6 3 : 3 3
− − − + = = =
6 6 = 3 = 3
( )
1 1 1 6 5 2 3 2 3 3 3 3
1 5 5 5 (^5 3 3 3 )
− − − −
− − − −
×
3 3 3 2 3 2 2 2
1 1 2
− − − − − −
− (^) −
×
9 5 2 2 2 5 1 2 2 1 1 2 2
− + − − +
− −
4 2 2 2
− − = = = =
( )
1 3 0
1 1 3 2 3 3
−
× ^ × − −
( )
1 (^3 3 )
1 1 1 1 3 3 3 3
− −
× − ×
0
3 2 (^3 32 )
2 2 1 3 3 2
a a a a
a a a a
3 2 2 3 3 2 2 1 2 3 3 2 2 1 3 2
a a
a
− − − +
−
4 4 4 2 2 2 3 3 3
3 2 , 0
a a a
a a
− − = = = =
