Question

progresie geometrica x+1/3,3x+1,5x+1​

Answer 1

Răspuns:

-1/6, -1/2, -3/2

Explicație pas cu pas:Trebuie satisfacuta, in acest caz, proprietatea

(3x+1)^2 = (x+1/3)(5x+1)

9x^2 + 6x + 1 = 5x^2 + x + 5x/3 + 1/3

4x^2 + 5x - 5x/3 ´+ 1 - 1/3 = 0

4x^2 + 5x(1-1/3) + 1-1/3 = 0

4x^2 + 5x * 2/3 + 2/3 = 0 I : 2/3(simplificam prin 2/3)

6x^2 + 5x + 1 = 0

x1,2 =(-5+-rad(25-24))/12 = (-5+-1)/12

x1= -6/12 = -1/2

x2 = -4/12 = -1/3

1) Pt x = -1/2 avem:

-1/2 + 1/3 = (-3+2)/6 = -1/6

3(-1/2) + 1  = -3/2 + 1 = -1si1/2 + 1 = -1/2

5(-1/2) + 1 = (-5 + 2)/2 = -3/2

Deci progresia geometrica:

-1/6, -1/2, -3/2

Verificare: (-1/6)*(-3/2) = 3/12 = 1/4 = (-1/2)^2 = 1/4, OK.

2) Pt x = -1/3 avem:

-1/3+1/3 = 0

-1 + 1 = 0

5(-1/3) + 1 = (-5+3)/3 = -2/3, care nu convine.

Concluzie: singura progresie geometrica cautata este cea de la punctul 1).