rozwiąż równanie: (x^2+3x) (x^2+4)(x^2+4x+4)=0
x²+3x=0 x²+4=0 x²+4x+4=0
x(x+3)=0 x²≠-4 Δ=16-16
x=0 x=-3 x₀=-4/2=-2
???
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x²+3x=0 x²+4=0 x²+4x+4=0
x(x+3)=0 x²≠-4 Δ=16-16
x=0 x=-3 x₀=-4/2=-2
???