Question

sa se determine x apartine R astfel incat tripletele de numere sa fie in progresie geometrica a) x+3\x-1 ; 2x+4\x-1; 4x+6\x-1

Answer 1

numerele sunt in progresie geometrica daca
primul inmultit cu ultimul este egal cu cel din mijloc la puterea a doua 

(x+3)/(x-1) ·(4x+6)/(x-1) = [(2x+4)/(x-1)]²
(x+3)(4x+6)/(x-1)²=(2x+4)²/(x-1)²
(4x²+18x+18)/(x-1)²=(4x²+16x+16)/(x-1)²
4x²+18x+18=4x²+16x+16
18x+18=16x+16
18x-16x=16-18
2x=-2
x=-1

a)(2x-1)/2·(2x+1) / (10x+1)=[(2x-1)/4]²
(2x-1)(2x+1)/2(10x+1)=(2x-1)²/16
(4x²-1)/2(10x+1)=(4x²-4x+1)/16
4x²-1=4x²-4x+1
-4x+1=-1
-4x=-2
x=1/2

b)(x+1)/3·(5x+1)=(3x-1)²
(x+1)(5x+1)/3=(3x-1)²
(5x²+x-5x+1)/3=(9x²-6x+1)
5x²-4x+1=9x²-6x+1
-4x²+2x=0
2x(-2x+1)=0
x=0
x=1/2