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

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Solution for 5=(5/(x^2+1))-1 equation: 


D( x )

x^2+1 = 0

x^2+1 = 0

x^2+1 = 0

1*x^2 = -1 // : 1

x^2 = -1

x in (-oo:+oo)

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

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

1-5*(x^2+1)^-1+5 = 0

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

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

1*(x^2+1)+5*(x^2+1)-5 = 0

x^2+5*x^2-4+5 = 0

6*x^2+1 = 0

(6*x^2+1)/(x^2+1) = 0

(6*x^2+1)/(x^2+1) = 0 // * x^2+1

6*x^2+1 = 0

6*x^2 = -1 // : 6

x^2 = -1/6

x belongs to the empty set

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