(x^-5y^4)^5/(2x^3y^-2)^2

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Solution for (x^-5y^4)^5/(2x^3y^-2)^2 equation: 


D( x )

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

x = 0

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

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

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

x^6 = 0

x = 0

x = 0

x = 0

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

((x^-5*y^4)^5)/((2*x^3*y^-2)^2) = 0

1/4*x^-31*y^24 = 0 // : 1/4*y^24

x^-31 = 0

x naleu017Cy do O

x belongs to the empty set

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