(4x^2y^0/8x^-2)^-2*(x^-2/x^-4y^5)

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Solution for (4x^2y^0/8x^-2)^-2*(x^-2/x^-4y^5) equation: 


D( x )

x = 0

x^-4 = 0

x^-2*((4*y^0*x^2)/8) = 0

x = 0

x = 0

x^-4 = 0

x^-4 = 0

1*x^-4 = 0 // : 1

x^-4 = 0

x naleu017Cy do O

x^-2*((4*y^0*x^2)/8) = 0

x^-2*((4*y^0*x^2)/8) = 0

x^-2*((4*x^2)/8) = 0

1/2 = 0

x belongs to the empty set

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

(x^-2*((4*y^0*x^2)/8))^-2*((x^-2)/(x^-4))*y^5 = 0

(x^-2*((4*x^2)/8))^-2*((x^-2)/(x^-4))*y^5 = 0

4*x^2*y^5 = 0 // : 4*y^5

x^2 = 0

x = 0

x in { 0} 

x belongs to the empty set

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