Question

(x-3)⁴-5(x-3)²+4=0
Помогите срочно
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Answer 1

$\displaystyle (x-3)^4-5(x-3)^2+4=0\\t^2-5t+4=0\\t=1,t=4\\(x-3)^2=1,(x-3)^2=4\\x_1=2,x_2=4,x_3=1,x_4=5$

Answer 2

Ответ:

t²-5t+4=0

t=1; t=4

(x-3)²=1

(x-3)²=4

x1=1; x2=2; x3=4; x4=5