Correct Answer: 0

Solution : Given:
$⇒x+ \frac{1}{x} =2$
By squaring both sides,
$({x}+ \frac{1}{x})^{2} =({2})^{2}$
$⇒{x}^{2}+ \frac{1}{x^{2}}+2×x×\frac{1}{x} =4$
$⇒{x}^{2}+ \frac{1}{x^{2}}=2$
Similarly, we can find,
$({x}+ \frac{1}{x})^{3} =({2})^{3}$
$⇒{x}^{3}+ \frac{1}{x^{3}}+3×x×\frac{1}{x}(x+ \frac{1}{x}) =8$
$⇒{x}^{3}+ \frac{1}{x^{3}}+3×2 =8$
$⇒{x}^{3}+ \frac{1}{x^{3}}=2$
Similarly, we can find the value of,
$⇒{x}^{99}+ \frac{1}{x^{99}}=2$
$\therefore {x}^{99}+ \frac{1}{x^{99}} -2=0$
Hence, the correct answer is 0.