rozwiaz [latex]3 x^{3} -10x^{3} +10x-3=0[/latex]

[latex]3x^4-10x^3+10x-3 = 0 \ 3(x^4-1)-10x(x^2-1) = 0 \ 3(x^2+1)(x^2-1)-10x(x^2-1) = 0 \ (3x^2+3-10x)(x^2-1) = 0 \ (3x^2-10x+3)(x+1)(x-1) = 0 \ Delta = (-10)^2-4*3*3 = 100-36= 64 \ sqrt{Delta} = 8 \ x_1 = frac{10-8}{2*3} = frac{1}{3} \ x_2 = frac{10+8}{2*3} = 3 \ 3(x-frac{1}{3})(x-3)(x+1)(x-1) = 0 \ (x-frac{1}{3})(x-3)(x+1)(x-1) = 0 \ x in {-1; frac{1}{3}; 1; 3 }[/latex]

3x^4-10x^³+10x-3=0 (3x^4-3)-(10x³-10x)=0 3(x^4-1)-10x(x²-1)=0 3(x²-1)(x²+1)-10x(x²-1)=0 (x-1)(x+1)(3x²-10x+3)=0 √Δ=100-4*3*3=√64=8 x1=10-86=13 x2=186=3 x=1   x=-1   x=13   x=3