Rozwiąż nierówność, potrzebuję to na jutro więc naj do pierwszej osoby. c) x^3-x^2-x+1[latex] leq [/latex]0 d) -x^4+10x^3+11x^2>0 e) 2x^3+x^2-8x-4>0 f) 3x^3-2x^2-6x+4[latex] leq [/latex]0

[latex]c) \x^3-x^2-x+1 leq 0\ x^2(x-1)-(x-1)=0\(x^2-1)(x-1)=0\(x+1)(x-1)(x-1)=0\x+1=0 vee x-1=0 vee x-1=0\x=-1 vee x=1 vee x=1\xin (-infty,-1>\\d)\ -x^4+10x^3+11x^2>0\-x^2(x^2+10x+11)>0\-x^2=0 vee x^2+10x+11=0\x_{1}=0\Delta=10^2-4cdot 11=56\x_{2}=frac{-10-2sqrt{14}}{2}=-5-sqrt{14}=-(5+sqrt{14})\x_{3}=frac{-10+2sqrt{14}}{2}=-5+2sqrt{14} = -(5-2sqrt{14})\xin(-(5-sqrt{14}), -(5+sqrt{14}))\ \ e)\ 2x^3+x^2-8x-4>0\x^2(2x+1)-4(2x+1)>0\(x^2-4)(2x+1)>0\(x+2)(x-2)(2x+1)>0\x+2=0 vee x-2=0 vee 2x+1=0\x=-2 vee x=2 vee x=-frac{1}{2}\xin(-2,-frac{1}{2})cup(2,+infty)\\f)\ 3x^3-2x^2-6x+4 leq 0\x^2(2x-2)-2(3x-2)le 0\(x^2-2)(3x-2)le 0\(x+sqrt{2})(x-sqrt{2})(3x-2)le0\x+sqrt{2}=0 vee x-sqrt{2}=0 vee 3x-2=0\x=-sqrt{2} vee x=sqrt{2} vee x=frac{2}{3}\xin(-infty,-sqrt{2}>cup[/latex]