решите уравнение (x+2)⁴+2x²+8x-16=0

(x + 2)^4 + 2(x^2 + 4x - 8) = 0 (x + 2)^4 + 2(x^2 + 4x + 4 - 12) = 0 (x + 2)^4 + 2((x + 2)^2 - 12) = 0 (x + 2)^4 + 2(x + 2)^2 - 24 = 0 Пусть (x + 2)^2 = t, тогдаt^2 + 2t - 24 = 0 D = 4 + 4*24 = 100 = 10^2t1 = ( - 2 + 10)/2 = 8/2 = 4;t2 = ( - 2 - 10)/2 = - 12/2 = - 6;1) (x + 2)^2 = 4(x + 2)^2 - 2^2 = 0(x + 2 - 2)(x + 2 + 2) = 0x (x + 4) = 0x1 = 0;x2 = - 42) (x + 2)^2 = - 6нет решенийОтвет- 4; 0