[tex]g)\\9x^{6}+6x^{5}+x^{4} = 0\\\\x^{4}(9x^{2}+6x+1) = 0\\\\x^{4}(3x+1)^{2} = 0\\\\x=0 \ \vee(3x+1)^{2} = 0\\\\x = 0 \ \vee \ 3x+1 = 0\\\\x = 0 \ \vee \ 3x= -1 \ \ /:3\\\\x = 0 \ \vee \ x = -\frac{1}{3}\\\\x \in \{-\frac{1}{3}, 0\}[/tex]
[tex]h)\\x^{5}+4x^{4}=12x^{3}\\\\x^{5}+4x^{4}-12x^{3} = 0\\\\x^{3}(x^{2}+4x-12) = 0\\\\x = 0\\\\lub\\\\x^{2}+4x-12 = 0\\\\\Delta = b^{2}-4ac = 4^{2}-4\cdot1\cdot(-12) = 16+48 = 64\\\\\sqrt{\Delta} = \sqrt{64} = 8\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} =\frac{-4-8}{2\cdot1} = \frac{-12}{2} = -6\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-4+8}{2} = \frac{4}{2} = 2\\\\x \in \{-6, 0, 2\}[/tex]
[tex]i)\\x^{3}+4x= -5x^{2}\\\\x^{3}+5x^{2}+4x = 0\\\\x(x^{2}+5x+4) = 0\\\\x = 0\\\\lub\\\\x^{2}+5x+4 = 0\\\\\Delta = 5^{2}-4\cdot4 = 25-16 = 9\\\\\sqrt{\Delta} = \sqrt{9} = 3\\\\x_1 = \frac{-5-3}{2} = \frac{-8}{2} = -4\\\\x_2 = \frac{-5+3}{2} = \frac{-2}{2} = -1\\\\x \in\{-4,-1,0\}[/tex]
[tex]j)\\-\frac{1}{2}x^{4} + x^{3} = \frac{1}{2}x^{2}\\\\-\frac{1}{2}x^{4}+x^{3}-\frac{1}{2}x^{2} = 0 \ \ |\cdot(-2)\\\\x^{4}-2x^{3}+x^{2} = 0\\\\x^{2}(x^{2}-2x+1) = 0\\\\x^{2}(x-1)^{2} = 0\\\\x^{2}(x-1) = 0\\\\x = 0 \ \vee \ x-1 = 0\\\\x = 0 \ \vee \ x = 1\\\\x \in \{0,1\}[/tex]
