Question

Решить уравнение: 40 Баллов!

(x-1)⁴-x²+2x-13=0​

Answer 1

Объяснение:

$(x-1)^4-x^2+2x-13=0\\(x-1)^4-x^2+2x-1-12=0\\(x-1)^4-(x^2-2x+1)-12=0\\(x-1)^4-(x-1)^2-12=0$

Пусть (x-1)^2=t≥0    ⇒

$t^2-t-12=0\\D=49;\sqrt{D}=7}.\\ t_1=(x-1)^2=-3\notin.\\t_2=(x-1)^2=4\\x^2-2x+1=4\\x^2-2x-3=0\\D=16;\sqrt{D}=4\\ x_1=3;x_2=-1.$

Ответ: x₁=3     x₂=-1.

Answer 2

$(x-1)^4-x^2+2x-13=0\\\\(x-1)^4-(x^2-2x+1)+1-13=0\\\\(x-1)^4-(x-1)^2-12=0\\\\t=(x-1)^2\geq 0\ \ ,\ \ \ \ \ t^2-t-12=0\ \ ,\ \ t_1=4\ ,\ t_2=-3<0\ \ (Viet)\\\\(x-1)^2=4\ \ \ \Rightarrow \ \ \ x-1=\pm 2\\\\x_1=2+1=3\ ,\ \ x_2=-2+1=-1\\\\Otvet:\ \ x_1=3\ ,\ x_2=-1\ .$