(2x/(x^2-16))=(2/(x-4))-(1/(x-4))

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Solution for (2x/(x^2-16))=(2/(x-4))-(1/(x-4)) equation: 


D( x )

x^2-16 = 0

x-4 = 0

x^2-16 = 0

x^2-16 = 0

1*x^2 = 16 // : 1

x^2 = 16

x^2 = 16 // ^ 1/2

abs(x) = 4

x = 4 or x = -4

x-4 = 0

x-4 = 0

x-4 = 0 // + 4

x = 4

x in (-oo:-4) U (-4:4) U (4:+oo)

(2*x)/(x^2-16) = 2/(x-4)-(1/(x-4)) // - 2/(x-4)-(1/(x-4))

(2*x)/(x^2-16)-(2/(x-4))+1/(x-4) = 0

(2*x)/(x^2-16)-2*(x-4)^-1+1/(x-4) = 0

(2*x)/(x^2-16)-2/(x-4)+1/(x-4) = 0

(2*x*(x-4))/((x^2-16)*(x-4))+(-2*(x^2-16))/((x^2-16)*(x-4))+(1*(x^2-16))/((x^2-16)*(x-4)) = 0

2*x*(x-4)-2*(x^2-16)+1*(x^2-16) = 0

x^2-8*x-16+32 = 0

x^2-8*x+16 = 0

x^2-8*x+16 = 0

x^2-8*x+16 = 0

DELTA = (-8)^2-(1*4*16)

DELTA = 0

x = 8/(1*2)

x = 4 or x = 4

(x-4)^2 = 0

((x-4)^2)/((x^2-16)*(x-4)) = 0

((x-4)^2)/((x^2-16)*(x-4)) = 0 // * (x^2-16)*(x-4)

(x-4)^2 = 0

x-4 = 0 // + 4

x = 4

x in { 4} 

x belongs to the empty set

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