((2x^3y^2)/(-x^2y^5))^-2

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Solution for ((2x^3y^2)/(-x^2y^5))^-2 equation: 


D( x )

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

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

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

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

2*x*y^-3 = 0 // : 2*y^-3

x = 0

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

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

x^2*y^5 = 0 // : y^5

x^2 = 0

x = 0

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

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

1/4*x^-2*y^6 = 0 // : 1/4*y^6

x^-2 = 0

x naleu017Cy do O

x belongs to the empty set

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