jmc

Rationalizing the denominator, we get $$\frac{1}{x-\sqrt{x^2-1}}=\frac{x+\sqrt{x^2-1}}{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}=\frac{x+\sqrt{x^2-1}}{x^2-(x^2-1)}=x+\sqrt{x^2-1}.$$Thus, $2x+2\sqrt{x^2-1}=20,$ so $x+\sqrt{x^2-1}=10.$ Then $\sqrt{x^2-1}=10-x.$ Squaring both sides, we get $$x^2-1=100-20x+x^2.$$Hence, $x=\frac{101}{20}.$

Similarly, $$\frac{1}{x^2+\sqrt{x^4-1}}=\frac{x^2-\sqrt{x^4-1}}{(x^2+\sqrt{x^4-1})(x^2-\sqrt{x^4-1})}=\frac{x^2-\sqrt{x^4-1}}{x^4-(x^4-1)}=x^2-\sqrt{x^4-1},$$so $$x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=2x^2=\boxed{\frac{10201}{200}}.$$