Question

Si: x + y = 2; x 2 + y 2 = 3; x > y Hallar: P = x – y + x 4 + y 4 - 8

Answer 1

Respuesta:

$P = \frac{2\sqrt{2} + 1}{2}$

Explicación paso a paso:

Si: x + y = 2; x² + y² = 3 ; x > y

binomio al cuadrado

(x+y)² = 2²

x²+2xy+y² = 4

x²+y² + 2xy = 4

3 + 2xy = 4

2xy = 1

xy = 1/2

identidad de legendre

(x+y)²+(x-y)² = 2(x²+y²)

2² + (x-y)² = 2(3)

4 + (x-y)² = 6

(x-y)² = 2

x-y = √2

binomio al cuadrado

(x²+y²)² = 3²

x⁴+2x²y²+y⁴ = 9

x⁴+y⁴+2(xy)² = 9

x⁴+y⁴+2(1/2)² = 9

x⁴+y⁴+2(1/4) = 9

x⁴+y⁴+1/2 = 9

x⁴+y⁴ = 9-1/2

x⁴+y⁴ = 17/2

Hallar:

P = x - y + x⁴ + y⁴ - 8

P = √2 + 17/2 - 8

P = 2√2/2 + 17/2 - 16/2

P = 2√2/2 + 1/2

$P = \frac{2\sqrt{2} + 1}{2}$