Question

решите уравнение (2-x)^2-4(3x+1)^2=0

Answer 1

(2-x)^2-(2(3x+1))^2=0
разность квадратов
(2-x-2(3x+1))(2-x+2(3x+1))=0
(2-x-6x-2)(2-x+6x+2)=0
x(5x+4)=0
x=0
5x=-4
x=-4/5=-0,8
Ответ: -0,8; 0

Answer 2

(2 – x)²– 4 • (3x + 1)²= 0
4 – 4x + x²– 4 • (9x²+ 6x + 1) = 0
4 – 4x + x²– 36x²– 24x – 4 = 0
– 28x – 35x²= 0
– 7x • (4 + 5x) = 0
x • (4 + 5x) = 0
x = 0
(4 + 5x) = 0
x = 0
x = – 4/5 = – 0,8