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