Rozwiązane

x³ - 3x² - 3x + 9 = 0



Odpowiedź :

[tex]\\x^3-3x^2-3x+9=0 \\x^2(x-3)-3(x-3)=0 \\(x-3)(x^2-3)=0 \\(x-3)(x+\sqrt3)(x-\sqrt3)=0 \\x-3=0\vee x+\sqrt3=0\vee x-\sqrt3=0 \\x\in\{-\sqrt3,\sqrt3,3\}[/tex]

 

Roma

[tex]x^3 - 3x^2 - 3x + 9 = 0 \\\ x^2 \cdot (x - 3) - 3 \cdot (x - 3) = 0 \\\ (x - 3)(x^2 - 3) = 0 \\\ (x - 3)(x^2 - (\sqrt{3})^2) = 0 \\\ (x-3)(x - \sqrt{3})(x + \sqrt{3}) = 0 \\\ x - 3 = 0 \ \vee \ x - \sqrt{3} = 0 \ \vee \ x + \sqrt{3} = 0 \\\ x=3 \ \vee \ x = \sqrt{3} \ \vee \ x =- \sqrt{3} \\\\ Zatem: \\\ x \in \{-\sqrt{3}; \ \sqrt{3}; \ 3 \}[/tex]