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q=1/2x
warunek rozwiazalnosci |q|<1 ∧ x≠0
|1/2x|<1⇒ -1<1/2·x<1
-2<x<2 ∧ x≠0
Suma szeregu S=a1/(1-q)=1/2·x/(1-1/2x)=x/(2-x)
wiec
1+1/x=x/(2-x) mnoze przez x(2-x)
x(2-x)+2-x=x²
2x-x²+2-x=x²
2x²-x-2=0
Δ=1+16=17
x1=(1-√17)/4≈-3/4
x2=(1+√17)/4≈5/4
x1,x2∈D
ODP
x1=(1-√17)/4
x2=(1+√17)/4