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[tex]\displaystyle\bf (x^4+x^2+1)(x^4+x^2+2)=12\\\\(x^4+x^2+1)(x^4+x^2+1+1)=12\\\\[/tex]
пусть:
[tex]x^4+x^2+1 = t\\t > 0\\[/tex]
[tex]\displaystyle\bf t(t+1)=12\\\\t(t-1)-12=0\\\\t^2+t-12=0\\\\t_1=-4 \ \ \ \ \ t_2 = 3[/tex]
[tex]\displaystyle \bf x^4+x^2+1=3\\\\x^4+x^2+1-3=0\\\\x^4+x^2-2=0\\\\[/tex]
[tex]x^2=a\geq 0[/tex]
[tex]\displaystyle\bf a^2+a-2=0\\\\a_1=-2 \ \ \ \ \ a_2=1\\\\[/tex]
[tex]\displaystyle\bf x^2=1\\\\x=\pm1[/tex]
ответ: 1 ; -1
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Answers & Comments
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[tex]\displaystyle\bf (x^4+x^2+1)(x^4+x^2+2)=12\\\\(x^4+x^2+1)(x^4+x^2+1+1)=12\\\\[/tex]
пусть:
[tex]x^4+x^2+1 = t\\t > 0\\[/tex]
[tex]\displaystyle\bf t(t+1)=12\\\\t(t-1)-12=0\\\\t^2+t-12=0\\\\t_1=-4 \ \ \ \ \ t_2 = 3[/tex]
[tex]\displaystyle \bf x^4+x^2+1=3\\\\x^4+x^2+1-3=0\\\\x^4+x^2-2=0\\\\[/tex]
пусть:
[tex]x^2=a\geq 0[/tex]
[tex]\displaystyle\bf a^2+a-2=0\\\\a_1=-2 \ \ \ \ \ a_2=1\\\\[/tex]
[tex]\displaystyle\bf x^2=1\\\\x=\pm1[/tex]
ответ: 1 ; -1