Resoluções das atividades de Matemática
Sumário
Capítulo 6 – Equação do 2o grau I ...1
Capítulo 7 – Equação do 2o grau II ...5
Capítulo 8 – Equação do 2o grau III ...7
Capítulo 9 – Sistemas de equações ...9
Capítulo 10 – Relação binária ...13
Definição – pág. 50
a) (– 2)2 – 4 · (– 2) –12 = 0
4 + 8 – 12 = 0 0 = 0
Sim, – 2 é solução da equação.
b) Sim, pois t substitui x, na equação ax2 + bx + = 0
c) Sim, pois se m = 0, tem-se 0 · x2 – x + 4 = 0 ⇒ – x + 4 = 0
gerando uma equação do 1o grau, que é da forma ax + b = 0.
d) Não, a forma reduzida é 3x2 – 8x – 11 = 0.
a) a = 6 b = 5 c = 8 b) a = 1 b = – 4 c = 0 c) a = 0,7 b = 0 c = –0,7 d) a = 1 3 b = 1 2 c = 1 5 03 x2 + 2x + 1 – (x2 + 3x – 2x – 6) = 9 x2 + 2x + 1 – (x2 + x – 6) = 9 x2 + 2x + 1 – x2 – x + 6 – 9 = 0 x – 2 = 0 x = 2
A equação não é do 2o grau.
01 02 a) x2 + 4x + 4 + x2 + 6x + 9 = 1 2x2 + 10x + 12 = 0 (:2) x2 + 5x + 6 = 0 a = 1, b = 5, c = 6 b) 3 1 2 6 12 6 2 3 3 12 0 2 3 9 0 2 2 2 (x ) x x x x x + − = − + + − = − + − = a = –2, b = 3, c = –9 c) 2t2 + 4t – 9t + 3t2 = t2 + 2t + t + 2 4t2 – 8t – 2 = 0 (:2) 2t2 – 4t – 1 = 0 a = 2, b = – 4, c = –1 d) 2m2 – 12m + 18 = 3(m –2)2 2m2 – 12m + 18 = 3m2 – 12m + 12 –m2 + 6 = 0 a = –1, b = 0, c = 6 a) Completa a = 8, b = – 3, c = 2 b) Completa a = 5, b = 5 , c = –1 c) Incompleta a = 4, b = 0, c = – 8 d) Completa 9x2 – x + 6 = 0 a = 9, b = –1, c = 6 e) Incompleta – 3x2 + x = 0 a = –3, b = 1, c = 0 f) Incompleta a = 6, b = 0, c = 0 04 05
a) k – 2 ≠ 0 k ≠ 2 b) k + 3 ≠ 0 k ≠ – 3 c) x2 – kx – 2x + 2k = 0 x2 +(–k –2)x + 2k = 0 –k – 2 ≠ 0 ou 2k ≠ 0 –k ≠ +2 ou k ≠ 0 k ≠ – 2 ou k ≠ 0 a) 2m – 18 ≠ 0 {m ∈ R/m ≠ 9} 2m ≠ 18 m ≠ 9 b) Resposta pessoal. Exemplos: m = 1 ou m = 2 –16x2 + 6x – 1 = 0 e –14x2 + 6x – 1 = 0 a) 3x – 4x2 + 12x –5 = –5 – 5x2 – 4x2 + 5x2 + 3x + 12x – 5 + 5 = 0 x2 + 15x = 0 b) x2 – 9 – 4x + 2x2 = 4x2 + 3x – 19 x2 + 2x2 – 4x2 – 4x – 3x – 9 + 19 = 0 – x2 – 7x + 10 = 0 c) 4x2 – 4x + 1 – 3x – 3x2 = 21 – 7x 4x2 – 3x2 – 4x – 3x + 7x + 1 – 21 = 0 x2 – 20 = 0 d) 3 2 5 4 2 1 24 6 2 5 24 6 15 16 16 4 12 30 2 2 2 2 2 ( − )− ( − ) = ⋅ −( ) − − + − = − − x x x x x x x x x 331 30 16 12 6 4 0 4 2 0 2 2 2 x x x x x x + + − + − = − + + =
Resolução de equações incompletas – pág. 54
a) x(x–15) = 0 x – 15 = 0 x = 0 ou x = 15 S={0; 15} b) 9x2 – 36 = 0 9x2 = 36 x2 = 36 9 x2 = 4 ⇒ x = ± 2 S={–2; +2} c) x2 = −14 7 x2 = – 2 S=∅ 06 07 08 01 d) y y y ou y y y y S − − = = − − = − = ⇒ = − = − = − 6 5 4 0 0 6 5 4 0 6 5 4 30 4 15 2 0 15 2 ; a) x2 – 2x + 1 + 4x2 – 1 = 0 5x2 – 2x = 0 x(5x – 2) = 0 x = 0 ou 5x – 2 = 0 x = 2 5 S= 0 2 5 ; b) 3(x2 – 4) = x2 – 8x + 16 + 8x 3x2 – 12 = x2 – 8x + 16 + 8x 2x2 – 28 = 0 x x S 2 28 2 14 14 14 14 = = = ± = −
{
;+}
c) 2 4 4 1 11 20 5 1 3 20 2 8 4 4 11 5 4 2 2 2 ( ) ( ) ( ) ( ) ( x x x x x x x x x x − − − + = − ⋅ − − − + + = − + 33 2 12 15 5 20 15 3 8 0 3 8 0 0 3 8 8 3 2 2 2 ) ( ) x x x x x x x x x ou x x − + = − + − + = − + = = − = − = SS= 0 8 3 ; d) 9 1 1 8 4 36 6 36 9 1 8 32 6 9 9 8 32 6 2 2 2 2 ( )( ) ( ) ( ) x x x x x x x x x x + − + − = − + − = − + − = −xx x b ac 2 2 2 32 3 0 4 32 4 1 3 1024 12 1036 − + = = − = − − ⋅ = + = ∆ ∆ ∆ ∆ ( ) ( ) ( ) 02x b a x x =− ± =− − ± ⋅ − = ± − ∆ 2 32 1036 2 1 32 2 259 2 ( ) ( ) S= − −
{
16 259;−16+ 259}
a) 2 3 3 1 1 3 1 3 1 3 2 2 6 3 3 x x x x x x x x x x x x x ( ) ( ) ( )( ) ( ) ( ) ( )( ) − − + + − = + − + − − − − = 22 2 3 3 7 0 − + − − = x x x x x = 0 ou x – 7 = 0 x = 7 S={0; 7} b) 9 2 6 6 12 6 9 6 12 2 12 9 6 2 2 2 2 2 x x x x x x x x x x x x x x − − − − = − − − − + = − + ( ) ( ) ( ) ( ) ( ) ++ − = − = − = = = = = 12 2 12 10 8 0 10 8 0 0 8 10 4 5 0 4 5 2 x x x x x x ou x x S ( ) ; c) 5 1 4 1 1 1 4 1 1 5 2 5 2 4 4 x x x x x x x x x x x x x ( ) ( )( ) ( )( ) ( )( ) ( + − − − + − = − + − + − − − + )) ( ) = − + − + − = − + = + = = + = = − 4 5 5 5 4 4 4 10 0 4 10 0 0 4 10 0 10 2 2 2 x x x x x x x x x ou x x 44 5 2 0 5 2 ⇒ = − = − x S ;Resolução de equações completas – pág. 56
a) x – 2 = ± 5 x = + 2 – 5 ou x = + 2 + 5 x = –3 ou x = 7 S = {–3; 7} 03 01 b) x + 3 = ± 7 x = – 3 – 7 ou x = – 3 + 7 x = –10 ou x = 4 S = {–10; 4} c) 8 – x = ± 1 – x = – 8 – 1 ou – x = – 8 + 1 – x = – 9 – x = – 7 x = 9 x = 7 S = {7; 9} d) 5x – 4 = ± 3 5x = + 4 – 3 ou 5x = 7 x= 1 5 ou x = 75 S= 1 5 7 5 ; e) 21 7 0 7 21 1 3 1 3 x x x S − = = = = 02 (x – 3)2 = 16 x – 3 = ± 4 x = +3 – 4 ou x = +3 + 4 x = –1 ou x = 7 S = {–1; 7} x – 3 = ± 6 x = 3 – 6 ou x = 3 + 6 x = – 3 ou x = 9 S = {–3; 9} x2 – 6x – 27 = 0 (– 3)2 – 6 · (– 3) – 27 = 0 9 + 18 – 27 = 0 – 3 é solução 92 – 6 · 9 – 27 = 0 81 – 54 – 27 = 0 9 é solução {–3; 9} a) Raízes racionais. b) Raízes irracionais. c) Raízes não reais. d) Raízes irracionais. e) Raízes não reais.
03
a) x2 – 18x + 81 = 100
(x – 9)2 = 100
Sim, a expressão que está no primeiro membro é um tri-nômio quadrado perfeito.
b) x – 9 = ±10 x = +9 – 10 ou x = +9 + 10 x = –1 x = 19 {–1; 19} a) x2 – 4x + 4 – 96 – 4 = 0 (x – 2)2 = 100 x – 2 = ± 10 x – 2 = –10 ou x – 2 = 10 x = – 8 x = 12 S = {– 8; 12} b) x2 – 2x + 1 – 48 – 1 = 0 (x – 1)2 = 49 x – 1 = 7 ou x – 1 = –7 x = 8 x = –6 S = {8; –6} c) x x x x ou x x 2 2 1 4 20 1 4 0 1 2 81 4 1 2 9 2 1 2 9 2 4 + + − − = + = + = + = − = x= 5 S= − −
{
4 5,}
d) 4 4 35 0 4 35 0 4 1 4 35 1 0 4 1 2 3 2 2 2 2 x x x x x x x − − = − − = − + − − = − = ( ) 66 1 2 36 4 1 2 9 1 2 3 1 2 3 7 2 2 2 x x x ou x x − = − = − = − = − = x=− = − 5 2 7 2 5 2 S , a) a = 4, b = 4, c = – 3 b) ∆ = b2 – 4ac = 42 – 4 · 4 · (– 3) ∆ = 16 + 48 ∆ = 64 05 124443 06 07c) Sim, pois o discriminante ∆ > 0.
d) x b a x x =− ± ⇒ − − ⋅ − + ⋅ =− − = − = − =− + = ∆ 2 4 64 2 4 4 64 2 4 4 8 8 12 8 3 2 4 8 8 1 2 44 8 1 2 3 2 1 2 = = − S ; a) ∆ = 32 – 4 · (1) · (– 28) ∆ = 9 + 112 ∆ = 121 x x x =− ± ⇒ = − − = − + 3 11 2 3 11 2 3 11 2 1 2 x1 = – 7 x2 = 4 S = {–7; 4} b) ∆ = 92 – 4 · (–1) · (–20) ∆ = 81 – 80 ∆ = 1 x x x =− ± − ⇒ =− − − =− + − 9 1 2 1 9 1 2 9 1 2 1 2 ( ) x1 = 5 x2 = 4 S = {4; 5} c) ∆ = (– 4)2 – 4 · (3) · 2 ∆ = 16 – 24 ∆ = –8 S = ∅ d) ∆ = (12)2 – 4 · 4 · 9 ∆ = 144 – 144 ∆ = 0 x b a x x S =− ± ⇒ = − − − = =− + − = = ∆ 2 3 0 2 4 3 8 3 0 2 4 3 8 3 8 1 2 ( ) ( ) a) x2 – 4x – 5 = 0 ∆ = (– 4)2 – 4 · 1 · (–5) ∆ = 16 + 20 ∆ = 36 x x x S = ± = = − = − = − = + = = = −
{
}
4 6 2 4 6 2 2 2 1 4 6 2 10 2 5 1 5 1 2 ; 08 09b) 9 – 12x + 4x2 – 24 + 4x + 3x + 9 = 0 4x2 – 5x – 6 = 0 ∆ = (–5)2 – 4 · (4) (–6) ∆ = 25 + 96 ∆ = 121 x b a x x x S =− ± =+ ± ⋅ = − = − = = = − ∆ 2 5 11 2 4 6 8 3 4 16 8 2 3 4 2 1 2 , c) 16 5 4 10 15 19 10 2 x − (x− )=− x+ 16x2 – 5x + 20 = –15x + 19 16x2 + 10x + 1 = 0 ∆ = (10)2 – 4 · (16) · 1 ∆ = 100 – 64 ⇒ ∆ = 36 d) 6 4 1 4 2 36 4 4 3 1 36 2 2 x x( − −) ( −x) = ( x + x+ ) 24x2 – 6x – 4(4 – 4x + x2) = 16x2 + 12x +4 24x2 – 16x2 – 6x – 12x – 4 – 16 + 16x – 4x2 = 0 4x2 – 2x – 20 = 0 : 2 2x2 – x – 10 = 0 ∆ = (–1)2 – 4(2) · (–10) ∆ = 1 + 80 ∆ = 81 x x x x S =− − ± ⋅ =+ ± = − = − = = = − ( ) ( ) ; 1 81 2 2 1 9 4 8 4 2 10 4 5 2 2 5 2 1 2 a) 6x ≠ 0 x ≠ 0 2 2 3 2 1 12 5 2 12 (x ) x x( ) ( ) x x x x − + − = + 2x – 4 + 6x2 – 3x = 5x2 + 2x x2 – 3x – 4 = 0 ∆ = (–3)2 – 4(+1) · (–4) ∆ = 9 + 16 ∆ = 25 x x x x x S =− ± ⋅
( )
= − = − = − = − = − − 10 6 2 16 16 32 4 32 1 2 1 8 1 2 1 8 1 2 1 2 ; 10 x b a x x x = − ± = + ± = − = − = = ∆ 2 3 5 2 2 2 1 8 2 4 1 2 S = {–1; 4} b) x – 3 ≠ 0 ⇒ x ≠ 3 {x ∈ R/x ≠ 3} 2 3 5 3 6 2 3 2 26 2 3 ( ) ( ) ( ) ( ) ( ) ( ) x x x x x x − + − + − = + − 6x – 10 + x2 + 6x – 3x – 18 = 2x + 52 x2 + 7x – 80 = 0 ∆ = 72 – 4 · 1 · (– 80) ∆ = 49 + 320 c) 1 3 3 3 3 3 3 3 3 2 ⋅ − − + + − = + − + − ( ) ( ) ( ) ( ) ( )( ) ( )( ) x x x x x x x x x – 3 – x2 – 6x – 9 = x2 – 9 –2x2 – 5x – 3 = 0 ∆ = (–5)2 – 4 (–2) · (–3) ∆ = 25 – 24 ∆ = 1 x b a x x x x x = − ± = − − ± −( )
⇒ = + − = − − = − = − ∆ 2 5 1 2 2 5 1 4 5 1 4 6 4 3 2 1 2 1 2 ( ) == − = − − 1 3 2 1 S ; x x x S = − ± =− + = − − =− − − + 7 3 41 2 7 3 41 2 7 3 41 2 7 3 41 2 7 3 41 2 1 2 ,Discussão das raízes – pág. 64
a) a = 1 b = 6m c = 9m2 – 4m – 8 b) (6m)2 – 4 · 1 · (9m2 – 4m – 8) 36m2 – 36m2 + 16m + 32 ∆ = 16m + 32 01
d) S b a P c a = − = − = = = − 0 36 0 1 36 a) x1 = –x2 ⇒ x1 + x2 = 0 simétricos − = ⇒ −b − = a m 0 3 9 1 0 –(3m – 9) = 0 3m – 9 = 0 ⇒ 3m = 9 ⇒ m = 3 b) x2 + 0x – 25 = 0 ⇒ x2 – 25 = 0 c) x = +5 e x = –5 S b a k k k k = − = ⇒ − − = ⇒ − = − = ⇒ = 2 1 3 2 1 2 6 2 5 5 2 ( ) P c a m m m m = ⇒ − = − ⇒ − = − − = − = − 2 1 2 3 5 2 1 2 15 2 1 30 29 2 x x x x c a p p p p p 1 2 1 2 1 1 1 1 3 6 1 1 1 3 6 1 1 3 6 3 5 5 3 = ⋅ = = ⇒ − = ⇒ − = − = − = = − x x x x b a x x x 2 1 1 2 1 1 1 3 4 8 1 4 8 2 = + =− =− − = = ( ) x x 2 2 3 2 6 = ⋅ = x x ca m m 1 2 12 1 12 ⋅ = = = a) S P b a c a m m m = − = − − = = = ( 3 ) 1 3 1 1 3 03 04 05 07 08 c) ∆ > 0 16m + 32 > 0 m> −32 16 ⇒ m > –2 d) ∆ = 0 16m + 32 = 0 m= −32 16 ⇒ m = –2 a) ∆ > 0 (8k – 3)2 – 4 · (1) · (16k2 – 15k) > 0 64k2 – 48k + 9 – 64k2 + 60k > 0 12k + 9 > 0 ⇒ k > − 9 12 ⇒ k > − 3 4 b) Resposta pessoal. Exemplos: p/k = 1 p/k = 0 c) x2 + 5x + 1 = 0 x2 – 3x + 0 = 0 x2 – 3x = 0 a) b² – 4ac < 0 (3)2 – 4 · (2) · [–(4 – 2m)] < 0 9 + 8 · (4 – 2m) < 0 9 + 32 – 16m < 0 –16m < –41 m> 41 16 b) b2 – 4ac < 0 (–4)2 – 4 · (5m – 2) · (1) < 0 16 – 20m + 8 < 0 –20m < –24 m>24 20 m>6 5
Relação entre os coeficientes e as raízes – pág. 65
a) S b a = − = − − − = − ( 17) 6 17 6 b) P c a = = − = − 3 6 1 2 a) x x S b a x x P c a 1 2 1 2 9 1 9 8 1 8 + = = − = − = − ⋅ = = = − = − b) S b a P c a = − = − − = = = − − = 1 9 1 9 8 9 8 9 c) S ba P c a = − =− − − = − = − = = − = ( 28) 7 28 7 4 0 7 0 02 03 01 02 06
Equações redutíveis à equação do 2o grau – Equações biquadradas – pág. 72 x2 = y y b a y y y =− ± =+ ± ⋅ ⇒ = = = = ∆ 2 9 7 2 4 16 8 2 2 8 1 4 1 2 4y2 – 9y + 2 = 0 ∆ = (–9)2 – 4 · 4 · 2 ∆ = 81 – 32 ∆ = 49 x x x x x 2 2 2 2 1 4 1 4 1 2 = ⇒ = ± = ⇒ = ± = ± S= − 2 + 2 −1 + 2 1 2 , , ,
02 x2 = y (para todos os itens):
a) y2 – 10y + 9 = 0 S = 10 y1 = 1 P = 9 y2 = 9 x x x 2 1 1 1 = = ± = ± x x x 2 9 9 3 = = ± = ± S = {–1; –3; +1; +3} b) 2y2 – 7y – 4 = 0 y y y = ± ⇒ = − = − = = 7 9 4 2 4 1 2 16 4 4 1 2 ∆ = (–7)2 – 4 · 2 · (–4) ∆ = 49 + 32 ∆ = 81 x x 2 1 2 = − ∃ ∈R x x x S 2 4 4 2 2 2 = = ± = ± = − +
{
;}
c) y2 – 12y = 0 y(y – 12) = 0 y = 0 e y – 12 = 0 ⇒ y = 12 x2 = y x x x 2 12 12 2 3 = = ± = ± x2 = 0 x = 0 S= −{
2 3 0; ;+2 3}
01 b) c) x2 = (x1)2 x 2 = 32 x2 = 9 x x c a x x x x 1 2 1 12 1 3 1 27 27 3 ⋅ = ⋅ = = = ( ) x1 + x2 = − b a 12 = –m m = –12Forma fatorada do trinômio do 2o grau – pág. 69
a) x2 – Sx + p = 0 d) x Sx p x x x x 2 2 2 0 7 5 2 5 0 5 7 2 0 − + = + + = + + = x2 – 8x + 15 = 0 b) x2 – Sx + p = 0 x2 + 4x – 21 = 0 c) x2 – Sx + p = 0 e) x Sx p x x 2 2 0 13 6 1 0 − + = − + = x2 + 3x + 2 = 0 a) x2− +(2 2)x+2 2 0= c) x2 – 2x – 1 = 0 b) x2 – 3 = 0 d) x b a ab x ab 2− + 1 0 + = 03 ∆ = 72 – 4 · 2 · 1 ∆ = 49 – 8 ∆ = 41 x x =− ± = − + 7 41 4 7 4 41 4 1 x2 7 4 41 4 = − −
Calculando-se o dobro das raízes: − +7 − − 2 41 2 7 2 41 2 e x Sx p x x x x 2 2 2 0 14 2 2 0 7 2 0 2 − + = + + = ⋅ + + = ( ) ⇒ 01 02 a) 5 13 6 5 40 60 5 2 3 5 5 2 6 5 3 5 6 2 2 x x x x x x x x x x − + − + = − ⋅ − − ⋅ − = − − ( ) ( ) ( ) == = − ⋅ − = − − 5 3 5 1 6 5 3 5 30 x x x x b) x x x x x x x x x 2 2 2 015 2 016 2 016 2 015 1 2 016 1 2 015 − − + + = + ⋅ − + ⋅ + = − ( ) ( ) ( ) ( ) 22 016 2 015 x+ 04
Capítulo 8
Equação do 2
ograu III
x x x x x x x x x x 1 2 1 2 1 2 1 2 1 2 8 2 2 3 1 16 2 3 1 2 2 + = ⋅ − − = + = − − = − − = − ( ) ⇒ 116 5 15 3 8 8 3 5 15 15 2 2 1 2 1 1 − − = = − = − = = = x x X x X X c a m ⇒ ⇒ +
d) 9y2 + 7y – 2 = 0 ∆ = b2 – 4ac ∆ = 49 – 4 · (9) · (–2) ∆ = 49 + 72 ∆ = 121 y y y = − ± = − = − = = 7 11 18 18 18 1 4 18 2 9 1 2 x x x x S 2 2 1 2 9 2 3 2 3 2 3 = − ∃ ∈ = = ± = − + R ; a) x + 1 ≠ 0 x – 1 ≠ 0 x ≠ – 1 x ≠ 1 U = {x ≠ 1 e x ≠ –1} ou U = R – {–1, 1} b) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) x x x x x x x x x x x x − ⋅ + + + + − = ⋅ − + − + 1 3 1 3 1 1 13 1 1 1 3 3 2 4 xx x x x x x x x x x x x 2 3 3 2 2 4 2 2 2 2 3 3 3 13 3 9 0 3 3 0 3 0 − − + + = − − = ⇒ ⋅ − = = ( ) oou x x x x S 2 2 3 0 0 3 0 3 0 3 − = = = ± = = −
{
, ,}
04 a) 24 + m · 22 + 36 = 0 16 + 4m + 36 = 0 4m + 52 = 0 m = −52 4 m = – 13 b) y4 – 13y2 + 36 = 0 x2 – 13x + 36 = 0 S = 13 x1 = 4 P = 36 x2 = 9 y2 = 4 y2 = 9 y = ±2 y = ±3 S = {–3; –2; +2; +3} a) x2 – 10x + 9 = 0 S = 10 x1 = 1 P = 9 x2 = 9 y2 = x y2 = 1 y = ±1 y2 = 9 y= ±3 S = {–1; –3; +1; +3} 03 05 b) Fazendo m2 = x 9x2 – 13x + 4 = 0 ∆ =169 – 4 · 9 · 4 ∆ =169 – 144 ∆ = 25 x x x = ± = = = = 13 5 18 8 18 4 9 18 18 1 1 1 m m S 2 2 4 9 2 3 1 1 2 3 1 2 3 1 = ⇒ ± = ⇒ ± = − − + ; ; ; c) a4 + 2a2 + 1 = 10 +1 – a2 a4 + 3a2 – 10 = 0 y2 + 3y – 10 = 0 S = 3 y1 = –5 P = –10 y2 = 2 a2 = y a2 = 2 a2 = –5 a = ± 2 ∃ ∈a R S={
− 2;+ 2}
Equações irracionais – pág. 73 x – 2 = 2x+20 Verificando: x2 – 4x + 4 = 2x + 20 − − = ⋅ − + − = 2 2 2 2 20 4 16 ( ) ( )F x2 – 6x – 16 = 0 S= = 6 x1 = –2 P = –16 x2 = 8 8 2 2 8 20 6 36 − = ⋅ + = ( )V S = {8} 02 ( 2x2+1)2= −(1 x)2 Verificando: 2x2 + 1 = 1 – 2x + x2 2 0 1 1 0 1 1 2 2 1 1 2 9 3 1 3 2 2 ⋅ + = − = − + = + = ={ }
( ) ( ) ( ) ( ) ; V V S x2 + 2x = 0 x(x+2) = 0 x = 0 e x + 2 = 0 x = –2 03 ( ) ( ) : ( ) ( ) 21 4 3 6 3 21 4 3 6 9 12 4 3 6 4 3 3 6 3 6 9 3 1 2 2 2 2 − − = − − = = − = − − = = x x x x x x 55 5 x= Verificando: 21 4 3 5 6 3 21 4 9 3 21 12 3 9 3 5 − ⋅ − = − = − = = ={ }
( )V S 01Sistema de equações – pág. 76 a) y = 3x – 9 x(3x – 9) = 12 3x2 – 9x – 12 = 0 ÷3 x2 – 3x – 4 = 0 S = 3 x1 = –1 P = – 4 x2 = +4 x = –1 x = 4 y = 3(–1) – 9 y = 3 · 4 – 9 y = –12 y = 3 (–1; –12) (4; 3) S = {(–1; –12), (4; 3)} b) (–1; –12) ⇒ ( ) ( ) ( ) ( ) ( ) − − − − = + ⋅ − = ⋅ − = − 1 12 13 12 1 12 13 12 13 13 12 1 12 2 (4; 3) ⇒ 4 3 13 3 16 3 13 3 13 13 3 1 3 2− ⋅ ⇒ − ⋅ = ⋅ = a) x = 1 – y (1 – y)2 – 2y2 = –14 1 – 2y + y2 – 2y2 = –14 –y2 – 2y + 15 = 0 y2 + 2y – 15 = 0 S = –2 y1 = –5 P = –15 y2 = 3 y = –5 y = 3 x = 1 –(–5) x = 1 – y x = 6 x = 1 – 3 (6; –5) x = –2 (–2; 3) S = {(6; –5), (–2; 3)} b) y = 2x + 3 x2 – (2x + 3)2 = –9 x2 – (4x2 + 12x + 9) = –9 x2 – 4x2 – 12x – 9 = – 9 –3x2 – 12x = 0 x (–3x – 12) = 0 x = 0 –3x – 12 = 0 –3x = 12 x = –4 x = 0 x = –4 y = 2 · 0 +3 y = 2x + 3 y = 3 y = 2 · (–4) + 3 (0; 3) y = – 8 + 3 y = – 5 (–4; – 5) S = {(0; 3), (– 4; –5)} 01 02 04 a) (4 2x−2)2= +(x 7)2 Verificando: 16 · (2x – 2) = x2 + 14x + 49 4 2 9 2 9 7 4 16 16 4 4 16 16 16 ⋅ − = + = ⋅ = = ( )V 32x – 32 = x2 + 14x + 49 x2 – 18x + 81 = 0 S = 18 x1 = 9 P = 81 x2 = 9 S = {9} b) ( ) ( ) ( ) ( ) ( ) 2 1 1 2 1 1 2 2 4 4 0 4 0 2 2 2 2 2 2 x x x x x x x x x x x x x + = + + = + + = = ⋅ − = − = Verificando: 2 0 1 1 0 1 1 2 4 1 1 4 9 1 2 3 3 ⋅ + = + = ⋅ + = + = + = ( ) ( ) V V x = 0 S = {0; 4} x – 4 = 0 ⇒ x = 4 c) x x x x x x + − = − + = + = = 1 1 1 1 1 1 0 2 2 ( ) ( ) Verificando: 0 1 0 1 0 1 1 0 + − = − = =
{ }
( )V S d) ( ) ( ) ( ) ( ) 6 1 3 6 1 9 1 3 1 9 8 2 2 2 2 + + = + + = + = + = = x x x x x Verificando: 6 8 1 3 6 9 3 6 3 3 9 3 8 + + = + = + = = ={ }
( )V S a) ( ( )) ( ) 2 3 4 2 3 256 3 128 125 4 4 4 x x x x + =( )
+ = + = = 2 128 4 256 4 2 4 2 4 125 4 4 8 4 2 ⋅ = = = = ={ }
( )V S b) 100 2 40 1000 100 2 50 10 2 40 100 100 100 10 40 50 10 3 3 ⋅ + = ⋅ = + = ⋅ = + = ( ) ( ) x x x 00 10 10 10 10 3 ⋅ = = = ( ) { } V x S Mergulhando Fundo – pág. 75 a) 1 1 2 1 2 1 3 1 3 1 4 1 4 1 5 1 5 1 6 − + − + − + − + − == = − =1 1 6 5 6 05 b) 1 1 2 1 2 1 3 1 6 1 7 1 999 1 1000 − + − + − + + − ... ... = = −1 1 = 1000 999 1000c) x = 4 –2y y2 – (4 –2y)y = 7 y2 – 4y + 2y2 = 7 3y2 – 4y – 7 = 0 ∆ = (– 4)2 – 4 · (3) · (–7) ∆ = 16 + 84 ∆ = 100 y y y =+ ± = = = − = − 4 10 6 14 6 7 3 6 6 1 1 2 y x x x x S = = − ⋅ = − = − = − − = − 7 3 4 2 7 3 4 14 3 12 14 3 2 3 2 3 7 3 2 3 7 3 ; ; ,, ( ;6 −1) y = –1 x = 4 – 2 · (–1) x = 4 + 2 x = 6 (6; –1) 2x2 + 3y = –13 2x – y = 3 2x – y = 3 2x – 3 = y 2x2 + 3(2x – 3) = –13 2x2 + 6x – 9 + 13 = 0 2x2 + 6x + 4 = 0 ( : 2) x2 + 3x + 2 = 0 ∆ = 9 – 8 ∆ = 1 x1 = –1 x = x2 = –2 para x1 = – 1 ⇒ y1 = – 5 para x2 = – 2 ⇒ y2 = – 7 S = {(–1, –5); (–2, –7)} a) 12x + 28 = x + 2y 12x + 28 = x + 2(14 –x2) 12x + 28 = x + 28 – 2x2 2x2 + 11x = 0 x(2x + 11) = 0 x = 0 e 2x = –11 x = −11 2 d) 123 – 3 ±1 2 03 x = 0 x y y y y = − = − − = − = − = − 11 2 14 11 2 14 121 4 56 121 4 65 4 2 y = 14 – 02 y = 14 (0; 14) S = ( ;0 14); 11; 2 65 4 − − b) xy= = − − = − ⋅ = 0 14 0 11 2 65 4 11 2 4 65 22 65 a) (y + 2)2 – y = 4 y2 + 4y + 4 – y = 4 y2 + 3y = 0 y = 0 ou y = –3 x = y + 2 x = y + 2 y = 0 y = –3 x = 0 + 2 x = –3 + 2 x = 2 x = –1 (2; 0) (–1; –3) S = {(2; 0), (–1; –3)} b) a = 5 + b (5 + b + b)2 = 9 5 + 2b = ±3 2b = –5 ± 3 2b = –5 –3 e 2b = –5 + 3 2b = – 8 2b = –2 b = – 4 b = –1 a = 1 a = 4 S = {(1; – 4), (4; –1)} c) 6 6 6 6 n m mn mn mn + = 6(m + n) = mn 6 · (–1) = mn ⇒ mn=–6 mn = –6 m = –3 m = 2 se ou m + n = –1 n = 2 n = –3 S = {(–3; 2), (2; –3)} d) 3 2 12 3 12 2 12 2 3 4 2 3 31 12 2 12 3 2 m n m n m n n n n n + = = − = − − + = − + + 22 19 2 = + n 04
ATIVIDADES DE MATEMÁTICA
ÁLGEBRA
S= – 6 n1 = –9 P = – 27 n2 = 3 m m m m = − ⋅ − = − ⋅ = + = = 4 2 9 3 4 2 3 3 4 6 10 2 ( ) S = {(10; –9), (2; 3)} a) 6 6 6 5 6 6 5 6 6 y x xy xy xy x y x y + = +(
)
= ⋅ ⇒ + = x + y = 5 x = 3 x = 2 xy = 6 ou y = 2 y = 3 S = {(3;2), (2;3)} b) x = 2y x=2(–1) 2y · y = 2 x = –2 y2 = 1 x = 2(+1) y = ± 1 x =2 S = {(2; 1), (–2; –1) c) 1 2 2 2 3 3 2 2 2 7 6 0 49 48 1 7 1 2 2 2 + = − = − − + = − − + − = = − = =− ± − = y x y x x x x x x x x ∆ ∆ ( ) −− ± − = − − = + = − − = 7 1 4 6 4 3 2 8 4 2 1 2 x x y = 2· 3 2 –3 y = 0 y = 2 · 2 – 3 y = 1 S= 3 2;0 , ( ; )2 1 d) 4x2 – 4xy + y2 + 4xy + y = 3 4x2 + y2 + y = 3 y = 2 – 2x 4x2+ 4 – 8x + 4x2 + 2 –2x = 3 8x2 – 10x + 3 = 0 ∆ = 100 – 96 ∆ = 4 x x y y x y y S = ± = ⇒ = − ⋅ ⇒ = = ⇒ = − ⋅ ⇒ = = 10 2 16 3 4 2 2 3 4 1 2 1 2 2 2 1 2 1 3 4 1 2 1 1 1 2 2 2 . , ; 1, . 2 1 3 12 2 12 2 3 4 2 3 31 12 2 12 3 2 m n m n n n n n = − = − − + = − + + 22 2 2 19 2 144 24 171 18 6 27 0 = + + + = + + − = n n n n n n 05 a) 2(y + 3) + 2(x + 4) = 60 2y + 6 + 2x + 8 = 60 2x + 2y = 46 x + y = 23 ⇒ x = 23 – y (x + 4)(y+3) = 126 (27 – y)(y + 3) = 126 27y – y2 + 81 – 3y = 126 –y2 + 24y – 45 = 0 ∆ = 576 – 4 · (–1) · (–45) ∆ = 576 – 180 ∆ = 396 y=− ± − − + 24 6 11 2 12 3 11 12 3 11 b) 2x + 2y = 360 x · y = 7 776 x + y = 180 x = 180 – y (180 – y) · y = 7 776 –y2 + 180y – 7 776 = 0 ∆ = 32 400 – 31 104 ∆ = 1 296 y y y =− ± − = − = = + = 180 36 2 90 18 72 90 18 108 1 2 x = 180 – 108 x = 180 – 72 x = 72 x = 108 Mergulhando Fundo – pág. 79Serão necessários cinco minutos para encher a banheira. Observe:
1. A torneira de água fria leva 400 s para encher a banheira, o que dá 1
400 da banheira em 1 s.
2. A torneira de água quente leva 480 s, o que dá 1 480da banheira em 1 s.
3. A água escoa da banheira em 800 s, o que dá 1 800 da banheira em 1 s. Então: 1 400 1 480 1 800 12 10 6 4 800 16 4 800 1 300 = − = + − = =
Isso é igual à quantidade de água acrescentada à banheira a cada segundo. Portanto, levará 300 segundos, ou 5 minu-tos, para encher a banheira.
Problemas do 2o grau – pág. 80 01 x2 + x = 6 x2 + x – 6 = 0 S = –1 x1 = –3 P = – 6 x2 = 2 {–3; 2} 02 3x2 – 2x = 21 3x2 – 2x – 21 = 0 ∆ =(–2)2 – 4 · (3)(–21) ∆ = 4 + 252 ∆ = 256 x b a x x x =− ± = + ± = + = + = − =− ∆ 2 2 16 6 18 6 3 14 6 7 3 1 2 (não satisfaz) {+3} 03 x2 – 5x = 50 x2 – 5x – 50 = 0 S = 5 x1 = +10 P = –50 x2 = –5 (não satisfaz) x = 10
Portanto, para o próximo quadrado perfeito, faltam 6 unidades.
04 (x + 2)2 = 100
x + 2 = ±10
x + 2 = –10 ou x + 2 = +10 x = –12 (não satisfaz) x = 8
O lado do quadrado inicial mede 8 cm. a) (6 + x)(4 + x) = 143 24 + 6x + 4x + x2 = 143 x2 + 10x – 119 = 0 ∆ = 102 – 4 · 1 · (–119) ∆ = 576 x x x =− ± = − = − = = 10 24 2 34 2 17 14 2 7 1 2 m (não satisfaz) b) 2(6 + 7) + 2 (4 + 7) 2 · 13 + 2 · 11 26 + 22 = 48 m 05 06 Consecutivos x – 1; x; x + 1 (x – 1)2 = (x + 1)2 – x2 x2 – 2x + 1 = x2 + 2x + 1 – x2 x2 – 4x = 0 x(x – 4) = 0 x = 0 x = 4 x – 1; x; x + 1 3; 4; 5 07 (x + 38) · (x + 20) = 1 600 x2 + 20x + 38x + 760 = 1 600 x2 + 58x – 840 = 0 ∆ = (58)2 – 4 · 1 · (– 840) ∆ = 3 364 + 3 360 ∆ = 6 724 x b a x x x = − ± ⇒ =− ± = − − = − = − + = ∆ 2 58 6 724 2 58 82 2 70 58 82 2 24 1 2 (n o conv m)ã é 22 =12 Resposta: 7
08 x ⇒ no de alunos y ⇒ no de balas por aluno
140 140 140 7 1 7 1 140 7 140 1 1 x y x y x y x y x x = ⇒ ⋅ = − = + ⇒ − + = − + = ( )( ) ( ) 440 140 980 7 140 980 7 0 7 980 0 49 4 1 980 4 2 2 + − − = − − = − − = − ⋅ − x x x x x x x x ∆ = ∆ = 99 3 920 3 969 7 63 2 35 28 1 2 + = + ± = = − ∆ = x x x (n o conv mã é ) Resposta: 35 alunos Mergulhando Fundo – pág. 81
Sejam a, b e c os lados de um triângulo; do enunciado, a + b + c = 12 => b + c = 12 – a. Uma condição necessária para um triângulo existir é que cada lado seja menor que a soma dos outros dois. Então a < b + c e, assim, a < 12 – a => 2ª < 12 => a < 6. As medidas dos lados de um triângulo são números positivos e, como são inteiros, os possíveis valores de a são 1, 2, 3, 4 e 5. Porém, o triângulo só vai existir se a = 2 => b + c = 10, com b = c = 5 ou a = 3 => b + c = 9, com b = 4 e c = 5 ou a = 4 => b + c = 8, com b = c = 4. Portanto, existem três triângulos com lados (2, 5, 5), (3, 4, 5) e (4, 4, 4).
Produto cartesiano – pág. 84 a) A × B = {(3; –2); (5; – 2)} b) A × C = {(3; 0); (3; – 4); (5; 0); (5; – 4)} c) C × B = {(0; –2); (– 4; –2)} d) B × B = {(–2; –2)} e) A × A = {(3; 3); (3; 5) (5; 3); (5; 5)} a) {(–4; –1), (0; –1), (4; –1)} b) {(–1; 7), (–1; 8)} c) {(–1; –1)} d) {(–4; 7), (–4; 8), (0; 7) (0; 8), (4; 7), (4; 8)} a) x = 7 d) x = –2 y = – 9 y = –2 b) x = 5 e) x = –1 y = – 4 y = 2 c) x = – 4 y = –3 a) 3 × 5 = 8 elementos b) 5 × 3 = 8 elementos c) Não. 05 B = {2, 4} A = {1, 3, 5} 06 4(x + 2) = 32 x + 2 = 8 x = 6 07 C A = {2, 3, 1} e B = {3, 4} 08 C 3 7 3 2 3 1 9 3 21 2 3 1 11 22 2 x y x y x y x y x x + ⋅ − ⇒ + − ⇒ = = = = = = ( ) 3 · 2 + y = 7 y = 7 – 6 y = 1 x2 + y – 1 22 + 1 – 1 = 4 09 Sendo m² = 16, m = ±4 e m +4 = 0 m = –4 Tem-se que m = –4 10 No ponto A: No ponto B: m2 – 2m – 15 = 0 m2 – 7m + 10 = 0 ∆ = 4 + 60 = 64 ∆ = 49 – 40 = 9 m =2 8 2 ± m =7 3 2 ± m = 5 ou m = –3 m = 5 ou m = 2 01 02 03 04 + 0 1 1 5 Quadrado de lado 5 d= 5 2 2 Plano cartesiano – pág. 82 01 8 5 7 2 1 0 6 34 8 9 5 7 2 1 0 3 4 6 C D F B E H G A J I 02 B C H G J A I E F D 03 A (3; 3) F (2; 0) B (–1; 3) G (0; 0) C (0; –3) H (0; –3) D (–2; –1) I (0; 2) E (1; –2) a) V f) V b) V g) V c) V h) V d) V i) V e) V v+( )vF⇒ 9+( )90= + =3 1 4 05 B A 0 C D Quadrado de lado 6 2p = 4 · 6 = 24 A = 62 = 36 06 04
b) B × A = {(–2,–2), (–2,–1), (–2,0), (–2,1), (–2,2), (–2,3), (0,–2), (0,–1), (0,0), (0,1), (0,2), (0,3), (4,–2), (4,–1), (4,0), (4,1), (4,2), (4,3)} c) {(x,y) ∈ A × B/x<y} {(–2,0), (–2,4), (–1,0), (–1,4), (0,4), (1,4), (2,4), (3,4)} 07 A = {–2, –1, 1, 2} B = {0, 1, 2, 3, 4} a) A × B = {(–2,0), (–2,1), (–2,2), (–2,3), (–2,4), (–1,0), (–1,1), (–1,2), (–1,–3), (–1,4), (1,0), (1,1), (1,2), (1,3), (1,4), (2,0), (2,1), (2,2), (2,3), (2,4)} d) {(x,y) ∈ B × A/x+y>0} {(–2,3), (0,1), (0,2), (0,3), (4,–2), (4,–1), (4,0), (4,1), (4,2), (4,3)} 4 3 2 1 2 1 –1 –2
Relação domínio-imagem e representação gráfica de uma relação – pág. 88 a) {(2,10), (4,8), (6,6), (8,4)} b) {(5,10), (6,10), (7,8), (7,10), (8,8), (8,10)} c) {(1,2), (1,4), (1,6), (2,2), (2,4), (3,2), (3,4), (4,2), (5,2)} a) R = {(3,5), (3,7), (3,9), (6,7), (6,9)} b) S = {(3,1), (3,3), (6,1), (9,1), (12,1)} c) T = {(3,1) (3,3) (3,5) (6,1) (6,3)} 03 R = {(– 4,–3), (–3,– 4) (3,4) (4,3)} ⇒ 4 pares 04 R = {(0,10); (1,9); (2,8); (3,7); (4,6); (5,5); (6,4)...,(10,0)} S = {(2,0); (3,1); (4,2); (5,3); (6,4); (7,5),...} R ∩ S = {(6,4)} 05 Gráfico de A Gráfico de B 4 3 2 1 3 2 1 06 A = {–2,–1,0,1,2,3} e B = {–2,0,4} a) A × B = {(–2,–2), (–2,0), (–2,4), (–1,–2), (–1,0), (–1,4), (0,–2), (0,0), (0,4), (1,–2), (1,0), (1,4), (2,–2), (2,0), (2,4), (3,–2), (3,0), (3,4)} 01 02 4 4 y x 1 3 3 2 2 1 –3 –2 –1 –1 –2 –3 –4 4 4 x y 3 3 2 2 1 1 –3 –4 –2 –1 –1 –2 –3 –4 4 4 y x 3 3 2 2 1 1 –3 –4 –2 –1 –1 –2 –3 –4 y x 3 2 1 –2 –1 –2 –1 4
y x 3 3 4 2 2 1 1 4 y x 2 2 1 1 –1 –1 y x 3 3 5 5 4 2 2 1 1 4 6 b) B × A 08 a) A × B d) {(x,y) ∈ B × A/ y = x 2} = {(2,1), (4,2)} c) y x 3 3 5 7 5 7 4 6 2 2 1 1 4 6 8 09 {(–2,0), (0,–2), (2,0), (0,2), (–1,–1), (–1,1), (1,–1), (1,1)} 10 Domínio ⇒ {–1, –2, –3, 0, 2} Contradomínio ⇒ {1, –3, 5, –7} A × B = {(–1,1), (–1,–3), (–1,5), (–1,–7), (–2,1), (–2,–3), (–2,5), (–2,–7), (–3,1), (–3,–3), (–3,5), (–3,–7), (0,1), (0,–3), (0,5), (0,–7), (2,1), (2,–3), (2,5), (2,–7)} Imagem ⇒ {(–1,1), (–1,5), (–2,1), (–2,3), (–2,5), (–2,–7), (–3,1), (–3,–3), (–3,5), (–3,–7), (0,1), (0,5)} y 2 –1 –2 2 –1 1 1 –2 x b) B × A = {(0,–2), (0,–1), (0,1), (0,2), (1,–2), (1,–1), (1,1), (1,2), (2,–2), (2,–1), (2,1), (2,2), (3,–2), (3,–1), (3,1), (3,2), (4,–2), (4,–1), (4,1), (4,2)} y x 3 3 4 2 2 1 1 –2 –1 –1 –2 –3 –4 4 c) {(x,y) ∈ A × B/ y = x2} = {(–2,4), (–1,1), (1,1), (2,4)} 4 4 3 3 2 1 2 1 –1 –1 –2 –2 –3 –3 –4 –4
