What is the answer to (x^2+1)/(x^2-4(x+1/x)+5)=0 ?

The answer to (x^2+1)/(x^2-4(x+1/x)+5)=0 is x=i,x=-i

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(x^2+1)/(x^2-4(x+1/x)+5)=0

\frac{x ^{2} + 1}{x ^{2} - 4 ( x + \frac{1}{x} ) + 5} = 0

x = i, x = - i

Solution steps

\frac{x ^{2} + 1}{x ^{2} - 4 ( x + \frac{1}{x} ) + 5} = 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{5 ^{2}}{x + 4 ^{2}} = 21

\frac{( x + 350 )}{x} + 315 = 92 x^{4} - 27

x, t \frac{1}{2} = \frac{0.693}{( x ^{2} \cdot 303 )}

\frac{1}{a} + \frac{1}{b} = \frac{1}{x}

\frac{x - 4}{x + 3} = 5

What is the answer to (x^2+1)/(x^2-4(x+1/x)+5)=0 ?
The answer to (x^2+1)/(x^2-4(x+1/x)+5)=0 is x=i,x=-i
Solve for x: (x^2+1)/(x^2-4(x+1/x)+5)=0
The solution is x=i,x=-i