1/(x^2-5x+6)-1/(x^2-3x+2)

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Solution for 1/(x^2-5x+6)-1/(x^2-3x+2) equation:


D( x )

x^2-(5*x)+6 = 0

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

x^2-(5*x)+6 = 0

x^2-(5*x)+6 = 0

x^2-5*x+6 = 0

x^2-5*x+6 = 0

DELTA = (-5)^2-(1*4*6)

DELTA = 1

DELTA > 0

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

x = 3 or x = 2

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

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

x^2-3*x+2 = 0

x^2-3*x+2 = 0

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

DELTA = 1

DELTA > 0

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

x = 2 or x = 1

x in (-oo:1) U (1:2) U (2:3) U (3:+oo)

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

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

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

x^2-5*x+6 = 0

x^2-5*x+6 = 0

x^2-5*x+6 = 0

DELTA = (-5)^2-(1*4*6)

DELTA = 1

DELTA > 0

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

x = 3 or x = 2

(x-2)*(x-3) = 0

x^2-3*x+2 = 0

x^2-3*x+2 = 0

x^2-3*x+2 = 0

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

DELTA = 1

DELTA > 0

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

x = 2 or x = 1

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

1/((x-2)*(x-3))-1/((x-1)*(x-2)) = 0

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

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

2 = 0

2/((x-2)*(x-3)*(x-1)) = 0

2/((x-2)*(x-3)*(x-1)) = 0 // * (x-2)*(x-3)*(x-1)

2 = 0

x belongs to the empty set

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