(x^3/4y^5/8)/(x^-1/4y^1/2)

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Solution for (x^3/4y^5/8)/(x^-1/4y^1/2) equation: 


D( x )

(((x^-1)/4)*y^1)/2 = 0

x = 0

(((x^-1)/4)*y^1)/2 = 0

(((x^-1)/4)*y^1)/2 = 0

(((x^-1)/4)*y)/2 = 0

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

x^-1 = 0

x naleu017Cy do O

x = 0

x = 0

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

((((x^3)/4)*y^5)/8)/((((x^-1)/4)*y^1)/2) = 0

((((x^3)/4)*y^5)/8)/((((x^-1)/4)*y)/2) = 0

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

x^4 = 0

x = 0

x in { 0} 

x belongs to the empty set

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