Odpowiedź:
[tex]zad.1~~x=-1~~\lor~~x=1~~\lor~~x=2\\\\zad.2~~x\in (-\infty ,0)\cup (\dfrac{3-\sqrt{5} }{2} ,\dfrac{3+\sqrt{5} }{2} )[/tex]
Szczegółowe wyjaśnienie:
[tex]zad.1\\\\-2x^{3} +4x^{2} +2x-4=0\\\\-2x^{2} \cdot (x -2)+2\cdot (x-2)=0\\\\(x-2)\cdot (-2x^{2} +2)=0\\\\(x-2)\cdot 2\cdot (1-x^{2} )=0~~\mid~~\div 2\\\\(x-2)\cdot (1-x^{2} )=0\\\\[/tex]
Korzystam ze wzoru skróconego mnożenia :
[tex]x^{2} -y^{2} =(x-y)\cdot (x+y)[/tex]
[tex]1 -x^{2} =1^{2} -x^{2} =(1-x)\cdot (1+x)\\\\(x-2)\cdot (1-x)\cdot (1+x)=0\\\\x-2=0~~\lor~~1-x=0~~\lor~~1+x=0\\\\x=2~~\lor~~x=1~~\lor~~x=-1[/tex]
Odp: Rozwiązanie równania : x = -1 ∨ x = 1 ∨ x = 2.
[tex]zad.2\\\\x^{3} -3x^{2} +x < 0\\\\x\cdot ( x^{2} -3x+1) < 0\\\\x=0~~\lor ~~x^{2} -3x+1=0\\\\x^{2} -3x+1=0~~~~\lor~~~~x=0\\\\a=1,~~b=-3,~~c=1\\\\\Delta=b^{2} -4ac\\\\\Delta=(-3)^{2} -4\cdot 1\cdot 1=9-4=5\\\\\sqrt{\Delta} =\sqrt{5} \\\\x_{1} =\dfrac{-b-\sqrt{\Delta} }{2a} ~~\lor~~x_{2} =\dfrac{-b+\sqrt{\Delta} }{2a}\\\\\\(~~x_{1} =\dfrac{3-\sqrt{5} }{2\cdot 1} ~~\lor~~x_{2} =\dfrac{3+\sqrt{5} }{2\cdot 1}~~)~~\land~~\sqrt{5} \approx 2,24\\\\x_{1} \approx 0,38~~\lor~~x_{2} \approx 2,62\\[/tex]
[tex]x=\dfrac{3-\sqrt{5} }{2} ~~\lor~~x=\dfrac{3+\sqrt{5} }{2} ~~\lor~~x=0\\\\\\x^{3} -3x^{2} +x < 0~~\Leftrightarrow~~x\in (-\infty ,0)\cup (\dfrac{3-\sqrt{5} }{2} ,\dfrac{3+\sqrt{5} }{2} )\\\\\\\\Odp:~~x\in (-\infty ,0)\cup (\dfrac{3-\sqrt{5} }{2} ,\dfrac{3+\sqrt{5} }{2} )[/tex]