Решить уравнение (x^2-3x)^2+2(x^2-3x)=0
X²-3x = t t² + 2t = 0 t(t + 2) = 0 t1 = 0; x²-3x = 0; x(x-3) = 0; x1 = 0; x2 = 3; t2 = -2; x² - 3x = -2; x²-3x + 2 = 0 => (x-1)(x-2) = 0; x3 = 1; x4 = 2 Ответ: x1 = 0; x2 = 3; x3 = 1; x4 = 2
(x² - 3x)² + 2(x² - 3 x ) = 0 ⇔ (x² - 3x) (x² - 3x + 2) = 0 ⇔ ⇔x(x - 3)(x² - 2x -x +2) = 0 ⇔ x(x - 3)[x(x-2) - (x-2)] = 0 ⇔ ⇔ x(x - 3)(x - 2)(x - 1) = 0 x ∈ {0, 1, 2, 3}