Question

Решите уравнение: (x-3)^2-(3x-2)^2=0

Answer 1

(x - 3 - (3x - 2)) * (x - 3 + (3x - 2)) = 0;
(x - 3 - 3x + 2) * ( x - 3 + 3x - 2) = 0 
( - 2x - 1) * ( 4x - 5) = 0 

- 2x - 1 = 0;
- 2x = 1
x = - 1/2 = - 0,5

4x - 5 = 0 
4x = 5
x = 5/4  = 1,25

Ответ:
- 0,5; 1.25 

Answer 2

$(x-3)^2-(3x-2)^2=0 \ x^{2} -6x+9-(9 x^{2} -12x+4)=0 \ x^{2} -6x+9-9 x^{2} +12x-4=0 \ -8 x^{2} +6x+5=0 \ 8 x^{2} -6x-5=0 \ D=36+160=196=14^2 \ x1=(6+14)/16=1,25 \ x2=(6-14)/16=-0,5 \ Otvet:1,25;-0,5$