(x^2+7)/(3x^2-5)=0

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Solution for (x^2+7)/(3x^2-5)=0 equation: 


D( x )

3*x^2-5 = 0

3*x^2-5 = 0

3*x^2-5 = 0

3*x^2 = 5 // : 3

x^2 = 5/3

x^2 = 5/3 // ^ 1/2

abs(x) = (5/3)^(1/2)

x = (5/3)^(1/2) or x = -(5/3)^(1/2)

x in (-oo:-(5/3)^(1/2)) U (-(5/3)^(1/2):(5/3)^(1/2)) U ((5/3)^(1/2):+oo)

(x^2+7)/(3*x^2-5) = 0

1*x^2 = -7 // : 1

x^2 = -7

x belongs to the empty set

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