1/(r+1)=4/(r+1)-6/(r^2-5r-6)

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Solution for 1/(r+1)=4/(r+1)-6/(r^2-5r-6) equation:


D( r )

r^2-(5*r)-6 = 0

r+1 = 0

r^2-(5*r)-6 = 0

r^2-(5*r)-6 = 0

r^2-5*r-6 = 0

r^2-5*r-6 = 0

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

DELTA = 49

DELTA > 0

r = (49^(1/2)+5)/(1*2) or r = (5-49^(1/2))/(1*2)

r = 6 or r = -1

r+1 = 0

r+1 = 0

r+1 = 0 // - 1

r = -1

r in (-oo:-1) U (-1:6) U (6:+oo)

1/(r+1) = 4/(r+1)-(6/(r^2-(5*r)-6)) // - 4/(r+1)-(6/(r^2-(5*r)-6))

1/(r+1)-(4/(r+1))+6/(r^2-(5*r)-6) = 0

1/(r+1)-4*(r+1)^-1+6/(r^2-5*r-6) = 0

1/(r+1)-4/(r+1)+6/(r^2-5*r-6) = 0

r^2-5*r-6 = 0

r^2-5*r-6 = 0

r^2-5*r-6 = 0

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

DELTA = 49

DELTA > 0

r = (49^(1/2)+5)/(1*2) or r = (5-49^(1/2))/(1*2)

r = 6 or r = -1

(r+1)*(r-6) = 0

1/(r+1)-4/(r+1)+6/((r+1)*(r-6)) = 0

1/(r+1)-4/(r+1)+6/(r+1) = 0

1-4+6 = 0

6-3 = 0

3 = 0

3/(r+1) = 0

3/(r+1) = 0 // * r+1

3 = 0

r belongs to the empty set

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