20/(x^2-2x)+5=10/(x-2)

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Solution for 20/(x^2-2x)+5=10/(x-2) equation: 


D( x )

x^2-(2*x) = 0

x-2 = 0

x^2-(2*x) = 0

x^2-(2*x) = 0

x^2-2*x = 0

x^2-2*x = 0

DELTA = (-2)^2-(0*1*4)

DELTA = 4

DELTA > 0

x = (4^(1/2)+2)/(1*2) or x = (2-4^(1/2))/(1*2)

x = 2 or x = 0

x-2 = 0

x-2 = 0

x-2 = 0 // + 2

x = 2

x in (-oo:0) U (0:2) U (2:+oo)

20/(x^2-(2*x))+5 = 10/(x-2) // - 10/(x-2)

20/(x^2-(2*x))-(10/(x-2))+5 = 0

20/(x^2-2*x)-10*(x-2)^-1+5 = 0

20/(x^2-2*x)-10/(x-2)+5 = 0

x^2-2*x = 0

x^2-2*x = 0

x*(x-2) = 0

x-2 = 0 // + 2

x = 2

x*(x-2) = 0

20/(x*(x-2))-10/(x-2)+5 = 0

20/(x*(x-2))+(-10*x)/(x*(x-2))+(5*x*(x-2))/(x*(x-2)) = 0

5*x*(x-2)-10*x+20 = 0

5*x^2-10*x-10*x+20 = 0

5*x^2-20*x+20 = 0

5*x^2-20*x+20 = 0

5*(x^2-4*x+4) = 0

x^2-4*x+4 = 0

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

DELTA = 0

x = 4/(1*2)

x = 2 or x = 2

5*(x-2)^2 = 0

(5*(x-2)^2)/(x*(x-2)) = 0

(5*(x-2)^2)/(x*(x-2)) = 0 // * x*(x-2)

5*(x-2)^2 = 0

x-2 = 0 // + 2

x = 2

x in { 2} 

x belongs to the empty set

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