2/x+2-3/x-2=-10/(x)2-4

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Solution for 2/x+2-3/x-2=-10/(x)2-4 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)

2/x-(3/x)-2+2 = -10/(x^2)-4 // + -10/(x^2)-4

2/x-(3/x)-(-10/(x^2))-2+2+4 = 0

2/x-3*x^-1+10*x^-2-2+2+4 = 0

10*x^-2-x^-1+4 = 0

t_1 = x^-1

10*t_1^2-1*t_1^1+4 = 0

10*t_1^2-t_1+4 = 0

DELTA = (-1)^2-(4*4*10)

DELTA = -159

DELTA < 0

x belongs to the empty set

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