((4)/(x^2-4x+3))/((x+1)/(x-1))

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Solution for ((4)/(x^2-4x+3))/((x+1)/(x-1)) equation: 


D( x )

x^2-(4*x)+3 = 0

(x+1)/(x-1) = 0

x-1 = 0

x^2-(4*x)+3 = 0

x^2-(4*x)+3 = 0

x^2-4*x+3 = 0

x^2-4*x+3 = 0

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

DELTA = 4

DELTA > 0

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

x = 3 or x = 1

(x+1)/(x-1) = 0

(x+1)/(x-1) = 0

x+1 = 0 // - 1

x = -1

x-1 = 0

x-1 = 0

x-1 = 0 // + 1

x = 1

x in (-oo:-1) U (-1:1) U (1:3) U (3:+oo)

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

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

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

x^2-4*x+3 = 0

x^2-4*x+3 = 0

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

DELTA = 4

DELTA > 0

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

x = 3 or x = 1

(x-1)*(x-3) = 0

4/((x-3)*(x+1)) = 0

x belongs to the empty set

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