(4x/x^2-25)=(20/x^2-25)-(3/x-5)

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Solution for (4x/x^2-25)=(20/x^2-25)-(3/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)

(4*x)/(x^2)-25 = 20/(x^2)-(3/x)-25+5 // - 20/(x^2)-(3/x)-25+5

(4*x)/(x^2)+3/x-(20/(x^2))-25-5+25 = 0

(4*x)/(x^2)+3/x-20*x^-2-25-5+25 = 0

7*x^-1-20*x^-2-5 = 0

t_1 = x^-1

7*t_1^1-20*t_1^2-5 = 0

7*t_1-20*t_1^2-5 = 0

DELTA = 7^2-(-20*(-5)*4)

DELTA = -351

DELTA < 0

x belongs to the empty set

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