Solution
Given, $$x+\ \frac{\ 1}{x}=2$$
$$x^2+1=2x^{ }$$
$$x^2-2x+1=0$$
$$\left(x-1\right)^2=0$$
$$x=1$$
Therefore, $$x^{100}+\ \frac{\ 1}{x^{100}}=1^{100}+\ \frac{\ 1}{1^{100}}$$
= 1 + 1
= 2
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