(40/1x^10)(3x^-12)(49/20x^2)

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Solution for (40/1x^10)(3x^-12)(49/20x^2) equation: 


D( x )

x = 0

x = 0

x = 0

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

3*(40/1)*(49/20)*x^-12*x^2*x^10 = 0

3*40*(49/20)*x^-12*x^2*x^10 = 0

294 = 0

x belongs to the empty set

x belongs to the empty set

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