((3x^3/2y^3)/(x^2y^-1/2))^-2

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Solution for ((3x^3/2y^3)/(x^2y^-1/2))^-2 equation: 


D( x )

(((3*x^3)/2)*y^3)/((x^2*y^-1)/2) = 0

(x^2*y^-1)/2 = 0

(((3*x^3)/2)*y^3)/((x^2*y^-1)/2) = 0

(((3*x^3)/2)*y^3)/((x^2*y^-1)/2) = 0

3*x*y^4 = 0 // : 3*y^4

x = 0

(x^2*y^-1)/2 = 0

(x^2*y^-1)/2 = 0

1/2*x^2*y^-1 = 0 // : 1/2*y^-1

x^2 = 0

x = 0

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

((((3*x^3)/2)*y^3)/((x^2*y^-1)/2))^-2 = 0

1/9*x^-2*y^-8 = 0 // : 1/9*y^-8

x^-2 = 0

x naleu017Cy do O

x belongs to the empty set

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