6+(1/(u+1))=(2/(u+2))

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Solution for 6+(1/(u+1))=(2/(u+2)) equation: 


D( u )

u+1 = 0

u+2 = 0

u+1 = 0

u+1 = 0

u+1 = 0 // - 1

u = -1

u+2 = 0

u+2 = 0

u+2 = 0 // - 2

u = -2

u in (-oo:-2) U (-2:-1) U (-1:+oo)

1/(u+1)+6 = 2/(u+2) // - 2/(u+2)

1/(u+1)-(2/(u+2))+6 = 0

1/(u+1)-2*(u+2)^-1+6 = 0

1/(u+1)-2/(u+2)+6 = 0

(1*(u+2))/((u+1)*(u+2))+(-2*(u+1))/((u+1)*(u+2))+(6*(u+1)*(u+2))/((u+1)*(u+2)) = 0

1*(u+2)-2*(u+1)+6*(u+1)*(u+2) = 0

6*u^2-u+18*u+12 = 0

6*u^2+17*u+12 = 0

6*u^2+17*u+12 = 0

6*u^2+17*u+12 = 0

DELTA = 17^2-(4*6*12)

DELTA = 1

DELTA > 0

u = (1^(1/2)-17)/(2*6) or u = (-1^(1/2)-17)/(2*6)

u = -4/3 or u = -3/2

(u+3/2)*(u+4/3) = 0

((u+3/2)*(u+4/3))/((u+1)*(u+2)) = 0

((u+3/2)*(u+4/3))/((u+1)*(u+2)) = 0 // * (u+1)*(u+2)

(u+3/2)*(u+4/3) = 0

( u+3/2 )

u+3/2 = 0 // - 3/2

u = -3/2

( u+4/3 )

u+4/3 = 0 // - 4/3

u = -4/3

u in { -3/2, -4/3 }

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