Question

X^4=(x-12)^2,помогите пж,

Answer 1

$x^4=(x-12)^2\\ x^2=\pm(x-12)\\ 1) \quad x^2-x+12=0; D=1-48=-47\ \textless \ 0$
$2)\quad x^2+x-12=0\to (x-3)(x+4)=0\to \left[\begin{matrix}x_1=3;\\x_2=-4.\end{matrix} \right.$