Răspuns:
Explicație pas cu pas:
b) log₂(1-x²)
1-x² > 0 => x² - 1 < 0 ; x² - 1 = 0 => x₁,₂ = ± 1
x I -∞ -1 1 +∞
x²-1 I ++++++++0---0++++++++ => x ∈ (-1 , 1)
c) log₁₍₂ (1+x²) ; x²+1 > 0 (∀) x ∈ R
d) log₄(x²+x-2)
x²+x-2 > 0 ; x²+x-2 = 0 => x₁,₂ = [-1±√(1+8)]/2 = (-1±3)/2
x₁ = -2 ; x₂ = 1 =>
x I -∞ -2 1 +∞
x²+x-2 I +++++++++0---------0+++++++++ =>
x ∈ (-∞ , -2)∪ (1 , +∞)
f) log₅(x²-x+1)
x²-x+1 > 0 ; x²-x+1 =0 => x₁,₂ = [1±√(1-4)] ; Δ < 0 =>
x²-x+1 > 0 pentru (∀) x ∈ R
h) log₁₍₂ (log₃x)
{x > 0
{log₃x > 0 <=> log₃x > log₃1 => x > 1 => x ∈ (1 , +∞)
