Question

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Answer 1

Explicación paso a paso:

9) Si: x + x⁻¹ =4. Hallar "x⁴ + x⁻⁴"

Resolvamos:

x + x⁻¹ = 4

(x + x⁻¹)² = (4)²

(x)² + 2(x)(x⁻¹) + (x⁻¹)² = 16

x² + 2(x⁰) + x⁻² = 16

x² + 2(1) + x⁻² = 16

x² + 2 + x⁻² = 16

x² + x⁻² = 16 - 2

x² + x⁻² =  14

Entonces:  

x² + x⁻² = 14

(x² + x⁻²)² = (14)²

(x²)² + 2(x²)(x⁻²) + (x⁻²)² = 196

x⁴ + 2(x⁰) + x⁻⁴ = 196

x⁴ + 2(1) + x⁻⁴ = 196

x⁴ + 2 + x⁻⁴ = 196

x⁴ + x⁻⁴ = 196 - 2

x⁴ + x⁻⁴ =  194

Por lo tanto, el valor de "x⁴ + x⁻⁴" es 194

10) Si: x+ y = 4: xy =-2. Hallar "x³ + y³"

Resolvamos:

x + y = 4

(x + y)³ = (4)³

x³ + 3xy(x + y) + y³ = 64

x³ + 3(-2)(4) + y³ = 64

x³ + -24 + y³ = 64

x³ + y³ = 64+24

x³ + y³ =  88

Por lo tanto, el valor de "x³ + y³" es  88

11) Si: x + y =2; xy =-3. Hallar "x³ + y³"

Resolvamos:

x + y = 2

(x + y)³ = (2)³

x³ + 3xy(x + y) + y³ = 8

x³ + 3(-3)(2) + y³ = 8

x³ + -18 + y³ = 8

x³ + y³ = 8+18

x³ + y³ =  26

Por lo tanto, el valor de "x³ + y³" es 26

12) Si: x + x⁻¹ = -2. Hallar "x⁶ + x⁻⁶"

Resolvamos:

x + x⁻¹ = -2

(x + x⁻¹)³ = (-2)³

(x)³ + 3(x)(x⁻¹)(x + x⁻¹) + (x⁻¹)³ = -8

x³ + 3(x⁰)(x + x⁻¹) + x⁻³ = -8

x³ + 3(1)(-2) + x⁻³ = -8

x³ -6 + x⁻³ = -8

x³ + x⁻³ = -8+6

x³ + x⁻³ =  -2

Entonces:

x³ + x⁻³ = -2

(x³ + x⁻³)² = (-2)²

(x³)² + 2(x³)(x⁻³) + (x⁻³)² = 4

x⁶ + 2(x⁰) + x⁻⁶ = 4

x⁶ + 2(1) + x⁻⁶ = 4

x⁶ + 2 + x⁻⁶ = 4

x⁶ + x⁻⁶ = 4 - 2

x⁶ + x⁻⁶ =  2

Por lo tanto, el valor de "x⁶ + x⁻⁶" es 2