(18)/(x^2-3x)-(6)/(x-3)=(5)/(x)

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Solution for (18)/(x^2-3x)-(6)/(x-3)=(5)/(x) equation:


D( x )

x^2-(3*x) = 0

x = 0

x-3 = 0

x^2-(3*x) = 0

x^2-(3*x) = 0

x^2-3*x = 0

x^2-3*x = 0

DELTA = (-3)^2-(0*1*4)

DELTA = 9

DELTA > 0

x = (9^(1/2)+3)/(1*2) or x = (3-9^(1/2))/(1*2)

x = 3 or x = 0

x = 0

x = 0

x-3 = 0

x-3 = 0

x-3 = 0 // + 3

x = 3

x in (-oo:0) U (0:3) U (3:+oo)

18/(x^2-(3*x))-(6/(x-3)) = 5/x // - 5/x

18/(x^2-(3*x))-(6/(x-3))-(5/x) = 0

18/(x^2-3*x)-6*(x-3)^-1-5*x^-1 = 0

18/(x^2-3*x)-6/(x-3)-5/x = 0

x^2-3*x = 0

x^2-3*x = 0

x*(x-3) = 0

x-3 = 0 // + 3

x = 3

x*(x-3) = 0

18/(x*(x-3))-6/(x-3)-5/x = 0

18/(x*(x-3))+(-6*x)/(x*(x-3))+(-5*(x-3))/(x*(x-3)) = 0

18-5*(x-3)-6*x = 0

15-6*x-5*x+18 = 0

33-11*x = 0

(33-11*x)/(x*(x-3)) = 0

(33-11*x)/(x*(x-3)) = 0 // * x*(x-3)

33-11*x = 0

33-11*x = 0 // - 33

-11*x = -33 // : -11

x = -33/(-11)

x = 3

x in { 3}

x belongs to the empty set

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