Ответ:
x^4-3x^2+2=0
a=x^2
a^2-3a+2=0
по теореме Виета
a1=2
a2=1
если a=x^2, то
x1=
x2=
x3=-1
x4=1
-3x^4+12x^2-12=0
-3a^2+12a-12=0
D=b^2-4ac=144-4*(-3)*(-12)=0
3x^4+9x^2-12=0
3a^2+9a-12=0
D=b^2-4ac=81-4*3*(-12)=225
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Ответ:
x^4-3x^2+2=0
a=x^2
a^2-3a+2=0
по теореме Виета
a1=2
a2=1
если a=x^2, то
x1=![\sqrt{2} \sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D)
x2=![-\sqrt{2} -\sqrt{2}](https://tex.z-dn.net/?f=-%5Csqrt%7B2%7D)
x3=-1
x4=1
-3x^4+12x^2-12=0
a=x^2
-3a^2+12a-12=0
D=b^2-4ac=144-4*(-3)*(-12)=0
x1=![-\sqrt{2} -\sqrt{2}](https://tex.z-dn.net/?f=-%5Csqrt%7B2%7D)
x2=![\sqrt{2} \sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D)
3x^4+9x^2-12=0
a=x^2
3a^2+9a-12=0
D=b^2-4ac=81-4*3*(-12)=225
x1=![-1 -1](https://tex.z-dn.net/?f=-1)
x2=![1 1](https://tex.z-dn.net/?f=1)