If $x+\frac {1}{x}=2,x\neq 0$$x^{99}+\frac {1}{x^{99}}=?$ Tuesday, March 29, 2022 We have, $x^2+1-2x=0$$\Rightarrow (x-1)^2=0$$\Rightarrow x=1$$\therefore x^{99}+\frac {1}{x^{99}}=1+1=2$

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