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

The answer to (x^2+1)/((x^2)-5(a+1/x)+2)=0 is x=i,x=-i;\quad x^2-5(a+1/x)+2\ne 0

Popular >

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

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

x = i, x = - i; x^{2} - 5 (a + \frac{1}{x}) + 2 \neq = 0

Solution steps

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

Simplify

\frac{x ^{2} + 1}{( x ^{2} ) - 5 ( a + \frac{1}{x} ) + 2} : \frac{x ^{2} + 1}{x ^{2} - 5 ( a + \frac{1}{x} ) + 2}

\frac{x ^{2} + 1}{x ^{2} - 5 ( a + \frac{1}{x} ) + 2} = 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; x^{2} - 5 (a + \frac{1}{x}) + 2 \neq = 0

\frac{\frac{22}{15}}{( \frac{1}{3} - 5 x )} = - \frac{22}{5}

\frac{7}{15 \cdot x + 3} = \frac{3}{5}

x, r + s - 2 t = (2 x + y)^{- 2}

\frac{x ^{1}}{( 243 ) x} = 3^{3}

\frac{x}{3 ( x + 1 )} = - 6

What is the answer to (x^2+1)/((x^2)-5(a+1/x)+2)=0 ?
The answer to (x^2+1)/((x^2)-5(a+1/x)+2)=0 is x=i,x=-i;\quad x^2-5(a+1/x)+2\ne 0