3/(x^2-36)=2/(x-6)-5/(X+6)

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Solution for 3/(x^2-36)=2/(x-6)-5/(X+6) equation: 


D( x )

x-6 = 0

X+6 = 0

x^2-36 = 0

x-6 = 0

x-6 = 0

x-6 = 0 // + 6

x = 6

X+6 = 0

X+6 = 0

X+6 = 0

x naleu017Cy do R

x^2-36 = 0

x^2-36 = 0

1*x^2 = 36 // : 1

x^2 = 36

x^2 = 36 // ^ 1/2

abs(x) = 6

x = 6 or x = -6

x belongs to the empty set

3/(x^2-36) = 2/(x-6)-(5/(X+6))

x belongs to the empty set

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