What is the answer to (x^2+1)/(x^2-1)=0 ?

The answer to (x^2+1)/(x^2-1)=0 is x=i,x=-i

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(x^2+1)/(x^2-1)=0

\frac{x ^{2} + 1}{x ^{2} - 1} = 0

x = i, x = - i

Solution steps

\frac{x ^{2} + 1}{x ^{2} - 1} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

x^{2} + 1 = 0

Solve

x^{2} + 1 = 0 : x = i, x = - i

x = i, x = - i

\frac{- 20}{x + 5} = \frac{- 10}{x}

x, \frac{a x + b}{c x + d} = 3

x, \frac{a x + b}{c x + d} = 9

\frac{x ^{2} - 2 x - 1}{( x - 1 ) ^{2}} = 0

\frac{2}{x + 3} = \frac{2 x - 3}{x}

What is the answer to (x^2+1)/(x^2-1)=0 ?
The answer to (x^2+1)/(x^2-1)=0 is x=i,x=-i
Solve for x: (x^2+1)/(x^2-1)=0
The solution is x=i,x=-i