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5x²+2x-1>0 ∧ x+2>0
√Δ=2√6
x1= (-√6-1)/5
x2= (-1+√6)/5
x nalezy (-∞ ; (-√6-1)/5)u((-1+√6)/5 , +∞)
x>-2
D x nalezy (-2,-√6-1)/5)u((-1+√6)/5 , +∞)
AD2
(2√2)^x=1/64
2^1*2^0.5x=2^-6
2^3/2x=2^-6
3/2x=-6
x=-4
Δ=4+20=24 ∧ x>-2
√Δ=√24=2√6
x₁=(2-2√6):10=(1-√6):5
x₂=(2+2√6):10=(1+√6):5
x∈(-∞;(1-√6):5)u((1+√6):5;∞) ∧x>-2→x∈((1+√6):5;∞)
zad.2 nie wiem czy dobrze rozszyfrowałam zapis
2²√2do potx=1/64
2dopot 2½x=2⁻⁶
2½x=-6
x=-6*2/5=-12/5