Question

Подскажите,пожалуйста,способы решения данного уравнения
(x^2-4x+5)^2=(x^2-2x-1)^2

Answer 1

$( {x}^{2} - 4x + 5) ^{2} = {( {x}^{2} - 2x - 1) }^{2} \\ ( {x}^{2} - 4x + 5) ^{2} - {( {x}^{2} - 2x - 1) }^{2} = 0\\ \\ (( {x}^{2} - 4x + 5) - ( {x}^{2} - 2x - 1)) (( {x}^{2} - 4x + 5) + ( {x}^{2} - 2x - 1)) = 0 \\ ( {x}^{2} - 4x + 5 - {x}^{2} + 2x + 1) ( {x}^{2} - 4x + 5 + {x}^{2} - 2x - 1) = 0 \\ ( - 2x + 6) ( 2{x}^{2} -6 x + 4 ) = 0 \\ - 2x + 6 = 0 \\ 2x = 6 \\ x = 3 \\ \\ 2 {x}^{2} - 6x + 4 = 0 \\ {x}^{2} - 3x + 2 = 0 \\ x_1x_2 = 2 \\ x_1 + x_2= 3 \\ x_1 = 1 \\ x_2 = 2$

Ответ: 1; 2; 3