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2010-03-26T15:11:36+01:00
A) 3(x²+4x+4)-2(x²+x-6)=8
3x²+12x+12-2x²-2x+12-8=0
x²+10x+16=0
∆=100-64
∆=36
√∆=6
x₁=(-10-6):2=-8 x₂=(-10+6):2=-2
c)x³+3x²-2x-6=0
x²(x+3)-2(x+3)=0
(x²-2)(x+3)=0
x²-2≠0
x²≠2
x≠√2 ∨ x≠√-2
x+3≠0
x≠-3
x∈R/{-3,√-2,√2}