((5x-10)/(x^2-4x+4))+((1)/(x-2))

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Solution for ((5x-10)/(x^2-4x+4))+((1)/(x-2)) equation: 


D( x )

x-2 = 0

x^2-(4*x)+4 = 0

x-2 = 0

x-2 = 0

x-2 = 0 // + 2

x = 2

x^2-(4*x)+4 = 0

x^2-(4*x)+4 = 0

x^2-4*x+4 = 0

x^2-4*x+4 = 0

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

DELTA = 0

x = 4/(1*2)

x = 2 or x = 2

x in (-oo:2) U (2:+oo)

(5*x-10)/(x^2-(4*x)+4)+1/(x-2) = 0

(5*x-10)/(x^2-4*x+4)+1/(x-2) = 0

x^2-4*x+4 = 0

x^2-4*x+4 = 0

x^2-4*x+4 = 0

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

DELTA = 0

x = 4/(1*2)

x = 2 or x = 2

(x-2)^2 = 0

(5*x-10)/((x-2)^2)+1/(x-2) = 0

(5*x-10)/((x-2)^2)+(1*(x-2))/((x-2)^2) = 0

1*(x-2)+5*x-10 = 0

6*x-12 = 0

(6*x-12)/((x-2)^2) = 0

(6*x-12)/((x-2)^2) = 0 // * (x-2)^2

6*x-12 = 0

6*x-12 = 0 // + 12

6*x = 12 // : 6

x = 12/6

x = 2

x in { 2} 

x belongs to the empty set

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