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2009-12-02T22:40:16+01:00
A) (3x+12)(x+2)(x²+1)³=0
(3x²+18x+24)(x⁶+1)=0
Δ=324-288=36 √Δ=6
x₁=-4 x₂=-2


b)x³-7x² +12x=0
x(x²-7x+12)=0
Δ=1
x₁=3 x₂=4
x(x-3)(x-4)=0
x=0 ∨ x-3=0 ∨ x-4=0
x=3 x=4

c) x³+3x²+6x+18=0
x²(x+3)+6(x+3)=0
(x+3)(x²+6)=0
x+3=0 ∨ x²+6=0
x=-3 x²=-6 x∈do zbioru pustego