5x^-20x/(x-4)^2

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Solution for 5x^-20x/(x-4)^2 equation: 


D( x )

(x-4)^2 = 0

x = 0

(x-4)^2 = 0

(x-4)^2 = 0

x-4 = 0 // + 4

x = 4

x = 0

x = 0

x in (-oo:0) U (0:4) U (4:+oo)

(5*x^-20*x)/((x-4)^2) = 0

(5*x^-19)/((x-4)^2) = 0

5*x^-19 = 0 // : 5

x^-19 = 0

x naleu017Cy do O

x belongs to the empty set

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