Question

(4x+1)^2totul supra 5=3x +7x-1 doar 7x-1 supra 3,3x sint intregi!Va rog ajutor dau coroana.
Rezolvati in R ecuatia

Answer 1

[(4x+1)^2]/5 = 3x+(7x-1)/3

(16x^2 +8x+1)/5 = (9x+7x-1)/3

(16x^2 +8x+1)/5 = (16x-1)/3

3(16x^2 +8x+1) = 5(16x-1)

48x^2 +24x+3=80x-5

48x^2 -56x +8=0, impartim prin 8

6x^2 -7x +1=0

6x^2 -6x-x+1=0

(6x^2 -6x)-(x-1)=0

6x(x-1)-(x-1)=0

(x-1)(6x-1)=0

x-1=0 sau 6x-1=0, adica:

x=1 sau x=1/6

Answer 2

(16x²+8x+1)/5=(16x-1)/3
48x²+24x+3=80x-5
48x²-56x+8=0  /:8
6x²-7x+1=0
6x²-6x-x+1=0
6x(x-1)-(x-1)=0
(x-1)(6x-1)=0
x-1=0⇒x=1
6x-1=0⇒x=1/6