97/(w+4)0=95/(w+4)0-(6/w)

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Solution for 97/(w+4)0=95/(w+4)0-(6/w) equation: 


D( w )

w = 0

(w+4)^0 = 0

w+4 = 0

w = 0

w = 0

(w+4)^0 = 0

t_1 = 1

t_1+0 = 0

t_1 = 0

t_1 = 0

w naleu017Cy do R

w+4 = 0

w+4 = 0

w+4 = 0 // - 4

w = -4

w belongs to the empty set

97/((w+4)^0) = 95/((w+4)^0)-(6/w)

w belongs to the empty set

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