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

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Solution for (4x-4)/(x^2-2x+1)=0 equation:


D( x )

x^2-(2*x)+1 = 0

x^2-(2*x)+1 = 0

x^2-(2*x)+1 = 0

x^2-2*x+1 = 0

x^2-2*x+1 = 0

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

DELTA = 0

x = 2/(1*2)

x = 1 or x = 1

x in (-oo:1) U (1:+oo)

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

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

x^2-2*x+1 = 0

x^2-2*x+1 = 0

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

DELTA = 0

x = 2/(1*2)

x = 1 or x = 1

(x-1)^2 = 0

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

4*x-4 = 0 // + 4

4*x = 4 // : 4

x = 4/4

x = 1

x in { 1}

x belongs to the empty set

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