Proszę o pomoc! Dwa zadania z wielomianów w załączniku na dziś!

[tex]1)\\\\\sqrt{3}x^3+3x^2-4\sqrt{3}x-12=0\\\\x^2(\sqrt{3}x+3)-4(\sqrt{3}x+3)=0\\\\(\sqrt{3}x+3)(x^2-4)=0\\\\(\sqrt{3}x+3)(x-2)(x+2)=0\\\\\sqrt{3}x+3=0\ \ \ \ \ \ \ \ \ \ \vee\ \ \ \ x-2=0\ \ \ \ \vee\ \ \ \ x+2=0\\\\\sqrt{3}x=-3\ \ |:\sqrt{3}\ \ \ \ \vee\ \ \ \ x=2\ \ \ \ \ \ \ \ \vee\ \ \ \ x=-2\\\\x=-\frac{3}{\sqrt{3}}\\\\x=-\frac{3}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\\\\x=-\frac{\not3\sqrt{3}}{\not3}\\\\x=-\sqrt{3}\ \ \ \ \vee\ \ \ \ x=2\ \ \ \ \vee\ \ \ \ x=-2[/tex]

[tex]2)\\\\\sqrt{2}x^3-2x^2+2022\sqrt{2}x-4044=0\\\\x^2(\sqrt{2}x-2)+2022(\sqrt{2}x-2)=0\\\\(\sqrt{2}x-2)(x^2+2022)=0\\\\\sqrt{2}x-2=0\ \ \ \ \ \ \ \ \vee\ \ \ \ x^2+2022=0\\\\\sqrt{2}x=2\ \ |:\sqrt{2}\ \ \ \ \vee\ \ \ \ x^2=-2022\ \ brak\ \ rozwiazania\\\\x=\frac{2}{\sqrt{2}}\\\\x=\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\\\\x=\frac{\not2\sqrt{2}}{\not2}\\\\x=\sqrt{2}[/tex]

Patro9kviz Patro9kviz · Answer 2

Odpowiedź:

a)

[tex]\sqrt{3}x^3+3x^2-4\sqrt{3}x-12=0\\\\ \sqrt{3}x^2(x+\sqrt{3})-4\sqrt{3}(x+\sqrt{3})=0\\\\[/tex]

[tex](x+\sqrt{3})(\sqrt{3}x^2-4\sqrt{3})=0[/tex] postać iloczynowa

1)

[tex]x+\sqrt{3}=0\\\\x=-\sqrt{3}[/tex]

2)

[tex]\sqrt{3}x^2-4\sqrt{3}=0\ \ \ /:\sqrt{3}\\\\ x^2-4=0\\\\x^2=4\ \ \ /\sqrt{}[/tex]

[tex]x=2[/tex] ∨ [tex]x=-2[/tex]

Rozwiązaniem równania są:

x∈{[tex]-2;-\sqrt{3};2[/tex]}

b)

[tex]\sqrt{2} x^3-2x^2+2022\sqrt{2}x-4044=0\\\\\sqrt{2}x^2(x-\sqrt{2})+2022\sqrt{2}( x-\sqrt{2})=0\\\\[/tex]

[tex](x-\sqrt{2})(\sqrt{2}x^2+2022\sqrt{2})=0[/tex] postać iloczynowa

1)

[tex]x-\sqrt{2}=0\\\\x=\sqrt{2}[/tex]

2)

[tex]\sqrt{2}x^2+2022\sqrt{2}=0\ \ \ \ /:\sqrt{2} \\\\x^2+2022=0\\\\x^2=-2002[/tex]

brak rozwiązań, ponieważ dowolna liczba podniesiona do kwadratu będzie nieujemna!

Rozwiązaniem równania jest:

x∈{[tex]\sqrt{2}[/tex]}