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D: x∈R
b)f(x)=1-x
D: x∈R
c)f(x)= 1/(x²+1)
x²+1≠0 stąd x²≠1 ponieważ x²≥0 ∀x∈R więc i tu:
D: x∈R
d)f(x)= 1/(x²-1)
x²-1≠0
x²≠1
x≠1 i x≠-1
D: x∈R\{-1,1}