Respuesta :

konrad509
[tex]D:x\not=0 \wedge 1+x^2\not=0\\ D:x\not =0\\\\ \dfrac{1}{x}-\dfrac{2x}{1+x^2}=0|\cdot x(1+x^2)\\ 1+x^2-2x^2=0\\ 1-x^2=0\\ x^2=1\\ x=-1 \vee x=1 [/tex]
(1/x) - (2x/(1+x^2)) = 0

Multiply by x(x^2+1)

1(x^2+1)+(−2x)(x)=0x(x2+1)

−x^2+1=0(Simplify both sides of the equation)

−x^2+1−1=0−1(Subtract 1 from both sides)

−x^2=−1

−x^2−1=−1−1(Divide both sides by -1)

x^2=1

x=ñ1(Take square root)

x=1 or x=−1