(1/x)-(1/(x-1))=((x-5)/3x)

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Solution for (1/x)-(1/(x-1))=((x-5)/3x) equation: 


D( x )

x = 0

x-1 = 0

x = 0

x = 0

x-1 = 0

x-1 = 0

x-1 = 0 // + 1

x = 1

x in (-oo:0) U (0:1) U (1:+oo)

t_1 = 0

t_1-(x-1)^-1-(x*((x-5)/3))+x^-1 = 0

(-1*x*(x-5))/3-1/(x-1)+t_1+1/x = 0

(-1*3*x)/(3*x*(x-1))+(-1*x^2*(x-5)*(x-1))/(3*x*(x-1))+(3*t_1*x*(x-1))/(3*x*(x-1))+(1*3*(x-1))/(3*x*(x-1)) = 0

3*t_1*x*(x-1)-1*x^2*(x-5)*(x-1)+1*3*(x-1)-1*3*x = 0

3*t_1*x^2-3*t_1*x-x^4+6*x^3-5*x^2-3*x+3*x-3 = 0

3*t_1*x^2-3*t_1*x-x^4+6*x^3-5*x^2-3 = 0

3*t_1*x^2-3*t_1*x-x^4+6*x^3-5*x^2-3 = 0

3*t_1*x^2-3*t_1*x-x^4+6*x^3-5*x^2-3

x^2*(3*t_1-x^2)-3*t_1*x+6*x^3-5*x^2-3

x^2*(6*x-5)+x^2*(3*t_1-x^2)-3*t_1*x-3

x^2*(6*x-5)-3*(t_1*x+1)+x^2*(3*t_1-x^2)

x^2*(3*t_1-x^2+6*x-5)-3*(t_1*x+1)

-2*(3*t_1-x^2+6*x-5)

(-2*(3*t_1-x^2+6*x-5))/(3*x*(x-1)) = 0

(-2*(3*t_1-x^2+6*x-5))/(3*x*(x-1)) = 0 // * 3*x*(x-1)

-2*(3*t_1-x^2+6*x-5) = 0

( -2 )

-2 = 0

x belongs to the empty set

( 3*t_1-x^2+6*x-5 )

3*t_1-x^2+6*x-5 = 0

DELTA = 6^2-(-1*4*(3*t_1-5))

DELTA = 4*(3*t_1-5)+36

4*(3*t_1-5)+36 = 0

4*(3*t_1-5)+36 = 0

12*t_1+16 = 0

12*t_1+16 = 0

12*t_1+16 = 0 // - 16

12*t_1 = -16 // : 12

t_1 = -16/12

t_1 = -4/3

DELTA = 0  <=>  t_3 = -4/3

x = -6/(-1*2) i t_1 = -4/3

x = 3 i t_1 = -4/3

( x = ((4*(3*t_1-5)+36)^(1/2)-6)/(-1*2) or x = (-(4*(3*t_1-5)+36)^(1/2)-6)/(-1*2) ) i t_1 > -4/3

( x = ((4*(3*t_1-5)+36)^(1/2)-6)/(-2) or x = ((4*(3*t_1-5)+36)^(1/2)+6)/2 ) i t_1 > -4/3

t_1+4/3 > 0

t_1+4/3 > 0 // - 4/3

t_1 > -4/3

x in { ((4*(0*3-5)+36)^(1/2)-6)/(-2)} 

x in { 3, ((4*(0*3-5)+36)^(1/2)+6)/2 }

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