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9-x²>0
-x²>-9
x²<9
x<3 ∧ x>-3 D=(-3,3)
b) [ x( x²-5)+(4x-1)(x-2) ] / (x-2)(x²-5) [ w w tej sytuacji mianownik musi być ≠0]
x-2≠0 ∨ x²-5≠0
x≠2 x²≠5
x≠√5 ∧ x≠-√5
D=R/ {-√5, 2, √5}
c)√2-3x≠0
-3x≠ -√2
3x≠√2 //3
x≠√2/3
D=R/{ √2/3}