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[tex]a)\\\\x^6+x=x+2x^4 \\\\ x^6+x-x-2x^4=0\\\\x^6-2x^4=0\\\\x^4(x^2-2)=0\\\\x^4(x-\sqrt{2})(x-\sqrt{2})=0\\\\x^4=0\ \ lub\ \ x-\sqrt{2}=0\ \ lub\ \ x+\sqrt{2}=0\\\\x=0\ \ lub\ \ x=\sqrt{2} \ \ lub\ \ x=-\sqrt{2}[/tex]

[tex]d)\\\\ \frac{x^4-4x^3}{2} =\frac{ x^5-6x^3 }{3}\ \ |*6\\\\3(x^4-4x^3)=2(x^5-6x^3)\\\\3x^4-12x^3 =2 x^5-12x^3\\\\3x^4-12x^3 -2 x^5+12x^3=0\\\\ -2x^5+3x^4\\\\x^4(-2x+3) =0\\\\ x^4=0\ \ lub\ \ -2x+3=0\\\\x=0\ \ lub\ \ -2x=-3\ \ :(-3)\\\\x=0\ \ lub\ \ x= \frac{3}{2}[/tex]

[tex]e)\\\\ \frac{1}{6}x^3+2x^2+6x=0\ \ |*6\\\\x^3+12x^2+36x=0\\\\ x(x^2+12x +36)x=0\\\\x(x+6)^2=0\\\\x=0\ \ lub\ \ x+6=0\\\\\ x=0\ \ lub\ \ x=-6[/tex]

[tex]f)\\\\ -4x^4+4x^3-x^2=0\\\\-x^2(4x^2-4x+1) =0\\\\-x^2(2x-1)^2=0\\\\-x^2=0\ \ lub\ \ 2x-1=0\\\\x=0\ \ lub\ \ 2x=1\ \ |:2\\\\x=0\ \ lub\ \ x=\frac{1}{2}[/tex]

Dorian112viz Dorian112viz · Answer 2

c)

[tex]x^{6} +x=x+2x^{4} \\x^{6} = 2x^{4} \\x^{6} -2x^{4} =0\\x^{4} (x^{2} -2)=0\\x=0\\x^{2} -2=0\\ x^{2} -2=(x-\sqrt{2} )(x+\sqrt{2} )\\ x = \sqrt{2} \\ x = -\sqrt{2}\\[/tex]

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

d)

[tex]\frac{x^{4} -4x^{3} }{2} = \frac{x^{5}-6x^{3} }{3} \\3x^{4} -12x^{3}=2x^{5}-12x^{3}\\3x^{4} =2x^{5} \\2x^{5}-3x^{4}=0\\x^{4}(2x-3)=0\\x=0\\2x-3=0\\x=1,5[/tex]

x={0,1.5}

e)

[tex]\frac{1}{6} x^{3} +2x^{2} +6x=0\\x(\frac{1}{6} x^{2} +2x+6)=0\\x = 0\\\frac{1}{6} x^{2} +2x+6=0\\[/tex]

[tex]x^{2} +12+36=0\\(x+6)^{2} =0\\x=-6[/tex]

x={-6,0}

f)

[tex]-4x^{4} +4x^{3} -x^{2} =0\\x^{2} (-4x^{2} +4x-1)=0\\-x^{2} (4x^{2}-4x+1 )=0\\-x^{2} (2x-1)^{2} =0\\[/tex]

x={0.5,0}