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$f(x)=\dfrac{a}{x}+\dfrac{b}{x-1}+\dfrac{c}{x+1}$<=>$a=-1,b=c=1$$f^{(2017)}(x)=\dfrac{2017!}{x^{2018}}-\dfrac{2017!}{(x-1)^{2018}}-\dfrac{2017!}{(x+1)^{2018}}$
$f(x)=\dfrac{a}{x}+\dfrac{b}{x-1}+\dfrac{c}{x+1}$<=>
ΑπάντησηΔιαγραφή$a=-1,b=c=1$
$f^{(2017)}(x)=\dfrac{2017!}{x^{2018}}-\dfrac{2017!}{(x-1)^{2018}}-\dfrac{2017!}{(x+1)^{2018}}$