Teste de Álgebra - Resolvido

Prévia do material em texto

TESTE DE VERIFICAÇÃO: ÁLGEBRA1 
 
1. Avalie cada expressão sem usar uma calculadora. 
 
(𝑎) (−3) = (−3) × (−3) × (−3) × (−3) = 9 × 9 = 81 
(𝑏) − 3 − (3 × 3 × 3 × 3) = −(9 × 9) = −81 
(𝑐) 3 =
1
3
=
1
3 × 3 × 3 × 3
=
1
9 × 9
=
1
81
 
(𝑑) 
5
5
= 5 = 5 = 5 × 5 = 25 
(𝑒) 
2
3
=
1
2
3
= 1 ÷
2
3
= 1 ×
3
2
=
9
4
 
(𝑓) 16 ⁄ = (4 × 4) ⁄ = (2 × 2 × 2 × 2) ⁄ = (2 ) ⁄ = 
= 2 ×( ⁄ ) = 2 =
1
2
=
1
2 × 2
=
1
4 × 2
=
1
8
 
 
2. Simplifique cada expressão. Escreva sua resposta sem expoentes 
negativos. 
 
(𝑎) √200 − √32 = 2 × 5 − 2 = (2 × 5 ) − (2 ) = 
= 2 × 5 − 2 = 2 × 5 − 2 = 2 × 2 × (5 − 2) == 2 × 3 × 2 = 
= 6√2 
(𝑏) (3𝑎 𝑏 )(4𝑎𝑏 ) = (3𝑎 𝑏 )[4 𝑎 (𝑏 ) ] = (3 × 16)(𝑎 × 𝑎 ) 
(𝑏 × 𝑏 ) = 48𝑎 𝑏 
(𝑐) 
3𝑥 𝑦
𝑥 𝑦
= 3𝑥 𝑦 =
1
3𝑥 𝑦
=
1
3 𝑥 𝑦
=
𝑥
9𝑦
 
 
1 Extraído dos Testes de Verificação, página XXI. 
3. Expanda e simplifique. 
 
(𝑎) 3(𝑥 + 6) + 4(2𝑥 − 5) = 3𝑥 + 18 + 8𝑥 − 20 = 11𝑥 − 2 
(𝑏) (𝑥 + 3)(4𝑥 − 5) = 4𝑥 − 5𝑥 + 12𝑥 − 15 = 4𝑥 + 7𝑥 − 15 
(𝑐) √𝑎 + √𝑏 √𝑎 − √𝑏 = √𝑎 − √𝑎 × √𝑏 + √𝑎 × √𝑏 − √𝑏 = 
= 𝑎 − √𝑎𝑏 + √𝑎𝑏 − 𝑏 = 𝑎 × − 𝑏 × = 𝑎 − 𝑏 
(𝑑) (2𝑥 + 3) = (2𝑥) + 2 × 2𝑥 × 3 + 3 = 4𝑥 + 12𝑥 + 9 
(𝑒) (𝑥 + 2) = 𝑥 + 3𝑥 × 2 + 3𝑥 × 2 + 2 = 𝑥 + 6𝑥 + 3𝑥 × 4 + 8 
= 𝑥 + 6𝑥 + 12𝑥 + 8 
 
4. Fatore cada expressão. 
 
(𝑎) 4𝑥 − 25 = (2𝑥) − 5 = (2𝑥 − 5)(2𝑥 + 5) 
(𝑏) 2𝑥 + 5𝑥 − 12 
Se 2𝑥 + 5𝑥 − 12 = 0, então: 
 ∆ = 𝑏 − 4𝑎𝑐 = 5 − 4 × 2 × (−12) = 25 + 96 = 121 
𝑥 =
−𝑏 ± √∆
2𝑎
=
−5 ± √121
2 × 2
=
−5 ± 11
4
 
𝑥 =
−5 + 11
4
=
6
4
=
3
2
⇒ 𝑥 −
3
2
= 0 ⇒ 2𝑥 − 2 ×
3
2
= 0 ⇒ 2𝑥 − 3 = 0 
𝑥 =
−5 − 11
4
= −
16
4
= −4 ⇒ 𝑥 + 4 = 0 
Logo 2𝑥 + 5𝑥 − 12 = (2𝑥 − 3)(𝑥 + 4) 
(𝑐) 𝑥 − 3𝑥 − 4𝑥 + 12 
Teorema do fator: Se 𝑃 é um polinômio e 𝑃(𝑏) = 0, então 𝑥 − 𝑏 é um 
fator de 𝑃(𝑥). Seja 𝑀(𝑎) o conjunto dos múltiplos de 𝑎: 
𝑀(12) = {−12, −6, −4, −3, −2, −1, 1, 2, 3, 4, 6, 12} 
𝑃(1) = 1 − 3 × 1 − 4 × 1 + 12 = 1 − 3 − 4 + 12 = 6 
𝑃(−1) = (−1) − 3 × (−1) − 4 × (−1) + 12 = −1 − 3 × 1 + 4 + 12 
= −1 − 3 + 16 = 16 − 4 = 12 
𝑃(2) = 2 − 3 × 2 − 4 × 2 + 12 = 8 − 3 × 4 − 8 + 12 = 8 − 12 + 4 = 
= 12 − 12 = 0 
 𝑥 − 3𝑥 − 4𝑥 + 12 𝑥 − 2 
 −𝑥 − 4𝑥 + 12 𝑥 − 𝑥 − 6 
 −6𝑥 + 12 
 (0) 
Se 𝑥 − 𝑥 − 6 = 0, então: 
∆ = 𝑏 − 4𝑎𝑐 = (−1) − 4 × 1 × (−6) = 1 + 24 = 25, 
𝑥 =
−𝑏 ± √∆
2𝑎
=
−(−1) ± √25
2 × 1
=
1 ± 5
2
 
𝑥 =
1 + 5
2
=
6
2
= 3 ⇒ 𝑥 − 3 = 0 
𝑥 =
1 − 5
2
= −
4
2
= −2 ⇒ 𝑥 + 2 = 0 
𝑥 − 3𝑥 − 4𝑥 + 12 = (𝑥 − 2)(𝑥 − 𝑥 − 6) = (𝑥 − 3)(𝑥 − 2)(𝑥 + 2) 
(𝑑) 𝑥 + 27𝑥 = 𝑥(𝑥 + 27) 
𝑀(27) = {−27, −9, −3, −1, 1, 3, 9, 27} 
𝑃(1) = 1 + 27 = 1 + 27 = 28 
𝑃(−1) = (−1) + 27 = −1 + 27 = −26 
𝑃(3) = 3 + 27 = 27 + 27 = 54 
𝑃(−3) = (−3) + 27 = −27 + 27 = 0 
𝑥 + 0𝑥 + 0𝑥 + 27 𝑥 + 3 
 −3𝑥 + 0𝑥 + 27 𝑥 − 3𝑥 + 9 
 9𝑥 + 27 
 (0) 
Se 𝑥 − 3𝑥 + 9 = 0, então: 
∆ = 𝑏 − 4𝑎𝑐 = (−3) − 4 × 1 × 9 = 9 − 36 = −27 < 0 
𝑥 + 27𝑥 = 𝑥(𝑥 + 27) = 𝑥(𝑥 + 3)(𝑥 − 3𝑥 + 9) 
(𝑒) 3𝑥 − 9𝑥 + 6𝑥 = 3𝑥 (𝑥 − 3𝑥 + 2) 
Se 𝑥 − 3𝑥 + 2 = 0, então: 
∆ = 𝑏 − 4𝑎𝑐 = (−3) − 4 × 1 × 2 = 9 − 8 = 1 
𝑥 =
−𝑏 ± √∆
2𝑎
=
−(−3) ± √1
2 × 1
=
3 ± 1
2
 
𝑥 =
3 + 1
2
=
4
2
= 2 ⇒ 𝑥 − 2 = 0 
𝑥 =
3 − 1
2
=
2
2
= 1 ⇒ 𝑥 − 1 = 0 
3𝑥 − 9𝑥 + 6𝑥 = 3𝑥 (𝑥 − 3𝑥 + 2) = 3𝑥 (𝑥 − 1)(𝑥 − 2) 
(𝑓) 𝑥 𝑦 − 4𝑥𝑦 = 𝑥𝑦(𝑥 − 4) = 𝑥𝑦(𝑥 − 2 ) = 𝑥𝑦(𝑥 − 2)(𝑥 + 2) 
 
5. Simplifique as expressões racionais. 
 
(𝑎) 
𝑥 + 3𝑥 + 2
𝑥 − 𝑥 − 2
 
Se 𝑥 + 3𝑥 + 2 = 0, então: Se 𝑥 − 𝑥 − 2 = 0, então: 
∆ = 3 − 4 × 2 = 9 − 8 = 1 ∆= (−1) − 4 × (−2) = 1 + 8 = 9 
𝑥 =
−3 ± √1
2 × 1
=
−3 ± 1
2
 𝑥 =
−(−1) ± √9
2 × 1
=
1 ± 3
2
 
𝑥 =
−3 + 1
2
= −
2
2
= −1 𝑥 =
1 + 3
2
=
4
2
= 2 
𝑥 =
−3 − 1
2
= −
4
2
= −2 𝑥 =
1 − 3
2
= −
2
2
= −1 
𝑥 + 3𝑥 + 2
𝑥 − 𝑥 − 2
=
(𝑥 + 1)(𝑥 + 2)
(𝑥 − 2)(𝑥 + 1)
=
𝑥 + 2
𝑥 − 2
 
(𝑏) 
2𝑥 − 𝑥 − 1
𝑥 − 9
×
𝑥 + 3
2𝑥 + 1
 
Se 2𝑥 − 𝑥 − 1 = 0, então: 
∆= (−1) − 4 × 2 × (−1) = 1 + 8 = 9 
𝑥 =
−(−1) ± √9
2 × 2
=
1 ± 3
4
 
𝑥 =
1 + 3
4
=
4
4
= 1 ⇒ 𝑥 − 1 = 0 
𝑥 =
1 − 3
4
= −
2
4
= −
1
2
⇒ 𝑥 +
1
2
= 0 ⇒ 2𝑥 + 1 = 0 
𝑥 − 9 = 0 ⇒ 𝑥 = 9 ⇒ 𝑥 = ±√9 = ±3 
2𝑥 − 𝑥 − 1
𝑥 − 9
×
𝑥 + 3
2𝑥 + 1
=
(𝑥 − 1)(2𝑥 + 1)(𝑥 + 3)
(𝑥 + 3)(𝑥 − 3)(2𝑥 + 1)
=
𝑥 − 1
𝑥 − 3
 
(𝑐) 
𝑥
𝑥 − 4
−
𝑥 + 1
𝑥 + 2
=
𝑥
(𝑥 + 2)(𝑥 − 2)
−
𝑥 + 1
𝑥 + 2
==
𝑥 − (𝑥 + 1)(𝑥 − 2)
(𝑥 + 2)(𝑥 − 2)
 
=
𝑥 − 𝑥 + 2𝑥 − 𝑥 + 2
(𝑥 + 2)(𝑥 − 2)
=
𝑥 + 2
(𝑥 + 2)(𝑥 − 2)
=
1
𝑥 − 2
 
(𝑑) 
𝑦
𝑥
−
𝑥
𝑦
1
𝑦
−
1
𝑥
=
𝑦
𝑥𝑦
−
𝑥
𝑥𝑦
÷
𝑥
𝑥𝑦
−
𝑦
𝑥𝑦
=
𝑦 − 𝑥
𝑥𝑦
×
𝑥𝑦
𝑥 − 𝑦
= 
=
(𝑦 + 𝑥)(𝑦 − 𝑥)
𝑥 − 𝑦
= −
(𝑥 + 𝑦)(𝑥 − 𝑦)
𝑥 − 𝑦
= −(𝑥 + 𝑦) 
 
6. Racionalize a expressão e simplifique. 
 
(𝑎) 
√10
√5 − 2
=
√10
√5 − 2
×
√5 + 2
√5 + 2
=
√10 √5 + 2
√5 − 2
=
√10 × √5 + 2√10
5 − 4
= 
= √2 × 5 × 5 + 2√10 = 5√2 + 2√10 
(𝑏) 
√4 + ℎ − 2
ℎ
=
√4 + ℎ − 2
ℎ
×
√4 + ℎ + 2
√4 + ℎ + 2
=
√4 + ℎ − 2
ℎ √4 + ℎ + 2
= 
=
4 + ℎ − 4
ℎ √4 + ℎ + 2
=
ℎ
ℎ √4 + ℎ + 2
=
1
√4 + ℎ + 2
 
 
7. Reescreva, completando o quadrado. 
 
Em geral, nós temos: 
𝑎𝑥 + 𝑏𝑥 + 𝑐 = 𝑎 𝑥 +
𝑏
𝑎
𝑥 + 𝑐 = 𝑎 𝑥 +
𝑏
𝑎
𝑥 +
𝑏
2𝑎
−
𝑏
2𝑎
+ 𝑐 
= 𝑎 𝑥 +
𝑏
𝑎
𝑥 +
𝑏
2𝑎
− 𝑎
𝑏
2𝑎
+ 𝑐 = 𝑎 𝑥 + 2
𝑏
2𝑎
𝑥 +
𝑏
2𝑎
+ 𝑐 − 
−𝑎
𝑏
4𝑎
= 𝑎 𝑥 +
𝑏
2𝑎
+ 𝑐 −
𝑏
4𝑎
 
(𝑎) 𝑥 + 𝑥 + 1 = 1 𝑥 +
1
2 × 1
+ 1 −
1
4 × 1
= 𝑥 +
1
2
+ 1 −
1
4
= 
= 𝑥 +
1
2
+
4
4
−
1
4
= 𝑥 +
1
2
+
3
4
 
(𝑏) 2𝑥 − 12𝑥 + 11 = 2 𝑥 −
12
2 × 2
+ 11 −
(−12)
4 × 2
= 
= 2 𝑥 −
12
4
+ 11 −
144
8
= 2(𝑥 − 3) + 11 − 18 = 2(𝑥 − 3) − 7 
 
8. Resolva a equação. (Encontre apenas as soluções reais). 
 
(𝑎) 𝑥 + 5 = 14 −
1
2
𝑥 ⇒ 2𝑥 + 2 × 5 = 2 × 14 − 2 ×
1
2
𝑥 ⇒ 
2𝑥 + 10 = 28 − 𝑥 ⇒ 2𝑥 + 𝑥 = 28 − 10 ⇒ 3𝑥 = 18 ⇒ 𝑥 = 18 ÷ 3 = 6 
𝑆 = {6} 
(𝑏)
2𝑥
𝑥 + 1
=
2𝑥 − 1
𝑥
⇒ 2𝑥 = (𝑥 + 1)(2𝑥 − 1) ⇒ 
2𝑥 = 2𝑥 − 𝑥 + 2𝑥 − 1 ⇒ 2𝑥 − 2𝑥 = 𝑥 − 1 ⇒ 𝑥 − 1 = 0 ⇒ 𝑥 = 1 
𝑆 = {1} 
(𝑐) 𝑥 − 𝑥 − 12 = 0 
∆ = (−1) − 4 × 1 × (−12) = 1 + 48 = 49 
𝑥 =
−(−1) ± √49
2 × 1
=
1 ± 7
2
 
𝑥 =
1 + 7
2
=
8
2
= 4 𝑒 𝑥 =
1 − 7
2
= −
6
2
= −3 
𝑆 = {−3, 4} 
(𝑑) 2𝑥 + 4𝑥 + 1 = 0 
∆ = 4 − 4 × 2 = 16 − 8 = 8 
𝑥 =
−4 ± √8
2 × 2
=
−4 ± √2
2 × 2
=
−4 ± 2√2
4
= −
4
4
±
2
4
√2 = −1 ±
√2
2
 
𝑆 = −1 −
√2
2
, −1 +
√2
2
 
(𝑒) 𝑥 − 3𝑥 + 2 = 0 
Se 𝑡 = 𝑥 , então: 
𝑥 − 3𝑥 + 2 = 0 ⇒ 𝑡 − 3𝑡 + 2 = 0 
∆ = (−3) − 4 × 1 × 2 = 9 − 8 = 1 
𝑡 =
−(−3) ± √1
2 × 1
=
3 ± 1
2
 
𝑡 =
3 + 1
2
=
4
2
= 2 ⇒ 𝑥 = 2 ⇒ ±√2 
𝑡 =
3 − 1
2
=
2
2
= 1 ⇒ 𝑥 = 1 ⇒ 𝑥 = ±1 
𝑆 = −√2, −1, 1, √2 
(𝑓) 3|𝑥 − 4| = 10 ⇒
3(𝑥 − 4) = 10
3[−(𝑥 − 4)] = 10
 
3(𝑥 − 4) = 10 ⇒ 3𝑥 − 12 = 10 ⇒ 3𝑥 = 10 + 12 ⇒ 𝑥 =
22
3
 
−3(𝑥 − 4) = 10 ⇒ −3𝑥 + 12 = 10 ⇒ 3𝑥 = 12 − 10 ⇒ 𝑥 =
2
3
 
𝑆 =
2
3
,
22
3
 
(𝑔) 2𝑥(4 − 𝑥) ⁄ − 3√4 − 𝑥 = 0 ⇒ 2𝑥(4 − 𝑥) ⁄ = 3√4 − 𝑥 ⇒ 
2𝑥(4 − 𝑥) ⁄ (4 − 𝑥) ⁄ = 3√4 − 𝑥 × √4 − 𝑥 ⇒ 
2𝑥(4 − 𝑥) = 3(4 − 𝑥) ⇒ 2𝑥 = 12 − 3𝑥 ⇒ 2𝑥 + 3𝑥 = 12 ⇒ 
5𝑥 = 12 ⇒ 𝑥 =
12
5
⇒ 𝑆 =
12
5
 
 
9. Resolva cada desigualdade. Escreva sua resposta usando a notação de 
intervalos. 
 
(𝑎) − 4 < 5 − 3𝑥 ≤ 17 
−4 < 5 − 3𝑥 ⇒ −4 − 5 < −3𝑥 ⇒ 3𝑥 < 9 ⇒ 𝑥 < 9 ÷ 3 = 3 
5 − 3𝑥 ≤ 17 ⇒ −3𝑥 ≤ 17 − 5 ⇒ 3𝑥 ≥ −12 ⇒ 𝑥 ≥ (−12) ÷ 3 = −4 
𝑆 = [−4, 3) 
(𝑏) 𝑥 < 2𝑥 + 8 
𝑥 − 2𝑥 − 8 < 0 ⇒ ∆ = (−2) − 4 × (−8) = 4 + 32 = 36 
𝑥 =
−(−2) ± √36
2 × 1
=
2 ± 6
2
 
𝑥 =
2 + 6
2
=
8
2
= 4 𝑒 𝑥 =
2 − 6
2
= −
4
2
= −2 
Definimos 𝑓(𝑥) = 𝑥 − 2𝑥 − 8, logo: 
𝑥 = −
𝑏
2𝑎
= −
−2
2 × 1
=
2
2
= 1 
𝑦 = 𝑓(𝑥 ) = 1 − 2 × 1 − 8 = 1 − 2 − 8 = 1 − 10 = −9 
Como 𝑥 < 𝑥 < 𝑥 e 𝑦 < 0, concluímos que 𝑆 = (−2, 4) 
(𝑐) 𝑥(𝑥 − 1)(𝑥 + 2) > 0 
Se 𝑥(𝑥 − 1)(𝑥 + 2) > 0, então: 
𝑥 = 0 ou 𝑥 − 1 = 0 ⇒ 𝑥 = 1 ou 𝑥 + 2 = 0 ⇒ 𝑥 = −2 
𝑥 = −3 ⇒ (−3)(−3 − 1)(−3 + 2) = (−3)(−4)(−1) = −12 < 0 
𝑥 = −1 ⇒ (−1)(−1 − 1)(−1 + 2) = (−1)(−2) × 1 = 2 > 0 
𝑥 = 0,5 ⇒ 0,5 × (0,5 − 1)(0,5+ 2) = 0,5 × (−0,5) × 2,5 = −0,625 < 0 
𝑥 = 2 ⇒ 2 × (2 − 1) × (2 + 2) = 2 × 1 × 4 = 8 > 0 
Logo 𝑆 = (−2, 0) ∪ (1, ∞) 
(𝑑) |𝑥 − 4| < 3 ⇒
𝑥 − 4 < 3
−(𝑥 − 4) < 3
 
𝑥 − 4 < 3 ⇒ 𝑥 < 3 + 4 ⇒ 𝑥 < 7 
−(𝑥 − 4) < 3 ⇒ 𝑥 − 4 > −3 ⇒ 𝑥 > 4 − 3 ⇒ 𝑥 > 1 
𝑆 = (1, 7) 
(𝑒) 
2𝑥 − 3
𝑥 + 1
≤ 1 ⇒
2𝑥 − 3
𝑥 + 1
× 𝑥 + 1 ≤ 𝑥 + 1 ⇒ 2𝑥 − 3 ≤ 𝑥 + 1 ⇒ 
2𝑥 − 𝑥 ≤ 1 + 3 ⇒ 𝑥 ≤ 4 
𝑥 + 1 ≠ 0 ⇒ 𝑥 ≠ −1 
𝑥 = 2 ⇒
2 × 2 − 3
2 + 1
=
4 − 3
2 + 1
=
1
3
< 1 
𝑥 = −2 ⇒
2 × (−2) − 3
(−2) + 1
=
−4 − 3
1 − 2
=
−7
−1
= 7 > 1 
𝑆 = (−1, 4] 
 
10. Diga se cada equação é verdadeira ou falsa. 
 
(𝑎) (𝑝 + 𝑞) = 𝑝 + 𝑞 Falso 
(𝑝 + 𝑞) = 𝑝 + 2𝑝𝑞 + 𝑞 
(𝑏) √𝑎𝑏 = √𝑎 × √𝑏 Verdadeiro 
√𝑎𝑏 = (𝑎𝑏) ⁄ = 𝑎 𝑏 = √𝑎√𝑏 
(c) √𝑎 + 𝑏 = 𝑎 + 𝑏 Falso 
(𝑑) 
1 + 𝑇𝐶
𝐶
= 1 + 𝑇 
1 + 𝑇𝐶
𝐶
=
1
𝐶
+
𝑇𝐶
𝐶
=
1
𝐶
+ 𝑇 
Falso 
(𝑒) 
1
𝑥 − 𝑦
=
1
𝑥
−
1
𝑦
 
1
𝑥
−
1
𝑦
=
𝑦
𝑥𝑦
−
𝑥
𝑥𝑦
=
𝑦 − 𝑥
𝑥𝑦
 
Falso 
(𝑓) 
1 𝑥⁄
𝑎 𝑥⁄ − 𝑏 𝑥⁄
=
1
𝑎 − 𝑏
 
1 𝑥⁄
𝑎 𝑥⁄ − 𝑏 𝑥⁄
=
1
𝑥
÷
𝑎 − 𝑏
𝑥
=
1
𝑥
×
𝑥
𝑎 − 𝑏
=
1
𝑎 − 𝑏
 
Verdadeiro 
 
REFERÊNCIAS: 
 
STEWART, James. Cálculo. Tradução: EZ2 Translate. São Paulo: Cengage 
Learning, 2013. v. 1.