`d)` $x^{200}$ `=x`
`⇔` $x^{200}$ `- x=0`
`⇔` $x^{200}$ `- x . 1=0`
`⇔` x . `($x^{199}$ - 1) =0`
⇔ \(\left[ \begin{array}{l}TH1: x^{199} - 1=0\\TH2: x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}TH1: x^{199} =1\\TH2: x =0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}TH1: x =1\\TH2: x =0\end{array} \right.\)
Vậy `x ∈ {0;1}`