a) x³-x²-x+1=0, rozwiązać rówanie b)x+6/x-2 = 2x+4/x-2 pomoże ktoś?:-)

a) x³-x²-x+1=0 x²(x-1)-(x-1)=0 (x²-1)(x-1)=0 (x-1)(x+1)(x-1)=0 x=1∨x=-1 b)x+6/x-2 = 2x+4/x-2 (x+6)(x-2)=(2x+4)(x-2) x²+4x-12=2x²-8 x²-4x+4=0 Δ=16-16=0 x=2

a) x³-x²-x+1=0 x²*(x-2) - 1*(x-1) = 0 x²-1 = 0 ∨ x-1=0 x² = 1 x = 1 x = ±1 b) x+6/x-2 = 2x+4/x-2 (x+6)(x+2) = (2x+4)(x-2) x²-2x+6x-12 = 2x²-4x+4x-8 x²-2x²-2x+6x+4x-4x-12+8 = 0 -x²+4x-4 = 0 a=-1 b=4 c=-4 Δ = b²-4ac Δ = 4²-4*(-1)*(-4) Δ = 16-16 Δ = 0 x = -b/2a x = -4/-2 x = 2