Question

Решите уравнение: (x+3)^4-(x+3)^2-12=0

Answer 1

$(x+3)^4-(x+3)^2-12=0\\t=(x+3)^2 textgreater 0; ,; ; t^4-t^2-12=0; ,; ; t_1=-3 textless 0,; t_2=4 textgreater 0\\(x+3)^2=4; ,; x+3=pm 2\\x_1=2-3=-1; ,; ; x_2=-2-3=-5$