Question

1. 12b-(9-7b+(5b-6)) = 12b-(9-7b+5b-6) = 12b-(3-2b) = 12b-3+2b =14b-3

2. 2b+4a) =3a-4-(5a-4b) = 3a-4-5a+4b =-2a+4b-4


3. (1+3x^4)^2 = (1+3x^4)

4. (x+4)(x-7) = x*x+x(4+7)+(7*4) =x^2+11x+28

5. (x^3+6)(x^3-8) = x^3*x^3-8x^3+6x^3-6*8 = x^6-2x^3-48

Answer 1

Resolver las siguientes igualdades.

1. 12b-(9-7b+(5b-6)) = 12b-(9-7b+5b-6) = 12b-(3-2b) = 12b-3+2b = 14b-3  

12b – (9 - 7b + 5b – 6) = 12b – 9 + 7b – 5b + 6 = 12b -3 +2b = 12b – 3 + 2b = 14b - 3

12b - 9 +7b - 5b + 6 = 14b – 3 =14b – 3 = 14b – 3 = 14b – 3

14b – 3 = 14b -3 = 14b – 3 = 14b – 3 = 14b – 3

0 = 14b – 3

14b = 3

b = 3/14

2. (2b+4a) =3a-4-(5a-4b) = 3a-4-5a+4b =-2a+4b-4  

2b + 4a = 3a – 4 – 5a + 4b = 3a – 4 – 5a + 4b = 2a + 4b - 4

2b + 4a = -2a + 4b – 4 = -2 a + 4b – 4 = 2a + 4b - 4

2b + 4a = 2a + 4b - 4

2b + 4a -2a -4b = - 4

2a – 2b = - 4

2(a – b) = - 4

a – b = - 4/2

a – b = - 2

3. (1+3x⁴)² = (1+3x⁴)  

1 + 6x⁴ + 9x⁸ = 1 + 3x⁴

6x⁴ + 9x⁸ - 3x⁴ = 1

3x⁴ + 9x⁸ = 1

9x⁸ + 3x⁴ = 1

4. (x+4)(x-7) = x*x+x(4+7)+(7*4) = x²+11x+28  

x² + 4x – 7x – 28 = x² + 4x + 7x + 28 = x² + 11x + 28

x² - 3x – 28 = x² + 11x + 28 = x² + 11x + 28

x² - 3x – 28 = 0

x = -(- 3) ± √[(- 3)² – 4(1)(- 28)] ÷ 2(1)

x = 3 ± √[9 + 112] ÷ 2

x = 3 ± √(121) ÷ 2

x = 3 ± 11 ÷ 2

x₁ = 3 + 11 ÷ 2  

x₁ = 14 ÷ 2

x₁ = 7

x₂ = 3 – 11 ÷ 2

x₂ = – 8 ÷ 2

x₂ = - 4

5. (x³+6)(x³-8) = x³*x³-8x³+6x³-6*8 = x⁶-2x³-48

x⁶ + 6x³ - 8x³ – 48 = x⁶ – 8x³ + 6x³ – 48 = x⁶ – 2x³ - 48

x⁶ - 2x³ – 48 = x⁶ – 2x³ – 48 = x⁶ – 2x³ - 48

x⁶ - 2x³ – 48  = 0