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Si F(x) = (x+20)/(x-1) , entonces:
F(F(x)) = F( (x+20)/(x-1)) = ( (x+20)/(x-1) + 20)/( (x+20)/(x-1) - 1)
F(F(x)) = [(x+20+20x-20)/(x-1)]/[(x+20-x+1)/(x-1)] = (21x)/(21)
F(F(x)) = x
Rpta: Alternativa C) x
• Ejercicio 5:
Si: P(x+1/x)=x³ + 1/x³ , entonces:
Para P(5) = P(x+1/x)=x³ + 1/x³
x + 1/x = 5
Si elevamos ambos miembros de la igualdad al cubo, obtendremos que:
(x+1/x)³ = 5
x³+3x²*1/x+3x*1/x² + 1/x³ = 125
x³+1/x³ + 3(x+1/x) = 125
Pero: P(5) = x³+1/x³ , y , x+1/x=5 , reemplazando:
P(5) + 3*5 = 125
P(5) = 125 - 15
P(5) = 110
Rpta: Alternativa A) 110
Saludos!! Jeizon1L
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