Odpowiedzi

2010-01-08T14:17:19+01:00
(2-x)(x²+3x+2)<0

Δ=9-8=1
√Δ=1
x₁=(-3-1)/2=-2
x₂=(-3+1)/2=-1

(2-x)(x+2)(x+1)<0
-(x-2)(x+2)(x+1)<0
(x-2)(x+2)(x+1)>0

x∈(-2,-1)u(2,+∞)

x³-3x²-5x+15≤0
x²(x-3)-5(x-3)≤0
(x²-5)(x-3)≤0
(x-√5)(x+√5)(x-3)≤0
x∈(-∞,-√5)u<√5, 3>

2010-01-08T14:18:31+01:00
(2-x)(x^2+3x+2)<0
x>2 x²+3x+2 a=1 b=3 c=2
Δ=b²-4ac=1 √Δ=1
x₁=(-b-√Δ)/2a x₂=(-b+√Δ)/2a
x₁=-2 x₂=-1


x³-3x²-5x+15≤0
x²(x-3)-5(x-3)≤0
(x²-5)(x-3)≤0
(x-√5)(x+√5)(x-3)≤0
x≤√5 ∨ x≤-√5 ∨ x≤3