Question

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Answer 1

Ответ:

$\mathbf{x_{1}=2};\\ \mathbf{ x_{2}=4};\\ \mathbf{x_{3}=0}.$

Пошаговое объяснение:

$(x-2)^4-4(x^2-4x+16)=0\\(x-2)^4-4(x-2)^2=0\\(x-2)^2=t\\t^2-4t=0; t(t-4)=0$

$t_{1}=0; t_{2}=4\\(x-2)^2=0; \mathbf{x_{1}=2};\\(x-2)^2=4; x-2=2;\mathbf{ x_{2}=4};x-2=-2;\mathbf{x_{3}=0}$