помогите пж там 3 примера. Created by Rahddc. matematika-ru

Ответ:x^4-3x^2+2=0a=x^2a^2-3a+2=0по теореме Виетаa1=2a2=1если a=x^2, тоx1=x2=x3=-1x4=1-3x^4+12x^2-12=0a=x^2-3a^2+12a-12=0D=b^2-4ac=144-4*(-3)*(-12)=0x1=x2=3x^4+9x^2-12=0a=x^23a^2+9a-12=0D=b^2-4ac=81-4*3*(-12)=225x1=x2=

помогите пж там 3 примера...

Rahddc @Rahddc

July 2022 1 4 Report

помогите пж там 3 примера

Answers & Comments

Maksik1241

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= $\sqrt{2}$

x2= $-\sqrt{2}$

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

$a=\frac{-b-\sqrt{D}}{2a}=$ $\frac{-12-0}{-6} =2$

x1= $-\sqrt{2}$

x2= $\sqrt{2}$

3x^4+9x^2-12=0

a=x^2

3a^2+9a-12=0

D=b^2-4ac=81-4*3*(-12)=225

$a=\frac{-b-\sqrt{D}}{2a}=$ $\frac{-9-15}{6}=\frac{-24}{6}=-4$

$a2=\frac{-b+\sqrt{D}}{2a}=$ $\frac{-9+15}{6}=\frac{6}{6}=1$

x1= $-1$

x2= $1$

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