(1)/(t-1)+(t)/(9t-6)=(1)/(9)

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Solution for (1)/(t-1)+(t)/(9t-6)=(1)/(9) equation:


D( t )

t-1 = 0

9*t-6 = 0

t-1 = 0

t-1 = 0

t-1 = 0 // + 1

t = 1

9*t-6 = 0

9*t-6 = 0

9*t-6 = 0 // + 6

9*t = 6 // : 9

t = 6/9

t = 2/3

t in (-oo:2/3) U (2/3:1) U (1:+oo)

1/(t-1)+t/(9*t-6) = 1/9 // - 1/9

1/(t-1)+t/(9*t-6)-(1/9) = 0

1/(t-1)+t/(9*t-6)-1/9 = 0

(1*9*(9*t-6))/(9*(t-1)*(9*t-6))+(9*t*(t-1))/(9*(t-1)*(9*t-6))+(-1*(t-1)*(9*t-6))/(9*(t-1)*(9*t-6)) = 0

1*9*(9*t-6)+9*t*(t-1)-1*(t-1)*(9*t-6) = 0

9*t^2-9*t^2+72*t+15*t-54-6 = 0

87*t-60 = 0

(87*t-60)/(9*(t-1)*(9*t-6)) = 0

(87*t-60)/(9*(t-1)*(9*t-6)) = 0 // * 9*(t-1)*(9*t-6)

87*t-60 = 0

87*t-60 = 0 // + 60

87*t = 60 // : 87

t = 60/87

t = 20/29

t = 20/29

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