x/x^2-25=5/x^2-25-4/x+5

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Solution for x/x^2-25=5/x^2-25-4/x+5 equation:


D( x )

x = 0

x^2 = 0

x = 0

x = 0

x^2 = 0

x^2 = 0

1*x^2 = 0 // : 1

x^2 = 0

x = 0

x in (-oo:0) U (0:+oo)

x/(x^2)-25 = 5/(x^2)-(4/x)-25+5 // - 5/(x^2)-(4/x)-25+5

x/(x^2)+4/x-(5/(x^2))-25-5+25 = 0

x/(x^2)+4/x-5*x^-2-25-5+25 = 0

5*x^-1-5*x^-2-5 = 0

t_1 = x^-1

5*t_1^1-5*t_1^2-5 = 0

5*t_1-5*t_1^2-5 = 0

DELTA = 5^2-(-5*(-5)*4)

DELTA = -75

DELTA < 0

x belongs to the empty set

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