(4x^3/7y^-2)^4/(2^-3x^-5y^4/5)^-1

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Solution for (4x^3/7y^-2)^4/(2^-3x^-5y^4/5)^-1 equation: 


D( x )

x = 0

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

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

x = 0

x = 0

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

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

1/40*x^-5*y^4 = 0 // : 1/40*y^4

x^-5 = 0

x należy do O

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

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

40*x^5*y^-4 = 0 // : 40*y^-4

x^5 = 0

x = 0

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

((((4*x^3)/7)*y^-2)^4)/(((2^-3*x^-5*y^4)/5)^-1) = 0

32/12005*x^7*y^-4 = 0 // : 32/12005*y^-4

x^7 = 0

x = 0

x in { 0} 

x belongs to the empty set

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