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f(x)=(x+4)/(x²+5x)
x²+5x≠0
x(x+5)≠0
x≠0 lub x+5≠
x≠-5
D:x∈R-{-5,0}
zad3
2⁻⁴/³*∛2⁵=2⁻⁴/³*2⁵/³=2⁻⁴/³⁺⁵/³=2¹/³=∛2
zad10
f(x)=(x+4)/x²-x
x²-x≠0
x(x-1)≠0
x≠0 i x-1≠0
x≠1
D:x∈R-{0,1}