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a)
f(x) = 5x/(x²-5x)
x²-5x ≠ 0
x(x-5) ≠ 0
x₁ ≠ 0
x-5 ≠ 0
x₂ ≠ 5
x ∈ (-∞;0)∪(0;5)∪(5;∞)
b)
f(x) = 5/(-x²-2x+3)
-x²-2x+3 ≠ 0
D = 4+16 = 16
x₁ =(-2-4)/2 =-3
x₂ =(-2+4)/2 = 1
(x+3)(x-1) ≠ 0
x₁ ≠-3
x₂ ≠ 1
x ∈ (-∞;-3)∪(-3;1)∪(1;∞)
c)
f(x) = (x²-4)/(x²+2x+2)
x²+2x+2 ≠ 0
D = 4-8 =-4 < 0 => x²+2x+2 ≠ 0zawsze !
x ∈ (-∞; ∞) lub x ∈ R
d)
f(x) = (x-4)/(x²+10x-6)
x²+10x-6 ≠ 0
D = 100+24 = 124
√D = √124 = 2√31
x₁ = (-10-2√31)/2 =-√31 -5
x₂ = (-10+2√31)/2 = √31 -5
(x+√31 +5)(x-√31 -5) ≠ 0
x₁ ≠ -√31 -5
x₂ ≠ √31 -5
x ∈ (-∞;-√31 -5)∪(-√31 -5;√31 -5)∪(√31 -5;∞)