toán 8 ( đại số)

N

noinhobinhyen

Do $a+b+c=0 \Rightarrow (a+b+c)^2=0 \Leftrightarrow a^2+b^2+c^2=-2(ab+bc+ac)$

$\Rightarrow ab+bc+ac = \dfrac{-1}{2}$

$\Rightarrow (ab+bc+ac)^2=\dfrac{1}{4}$

$\Leftrightarrow a^2b^2+b^2c^2+c^2a^c+2abc(a+b+c)=\dfrac{1}{4}$

$\Rightarrow 2(a^2b^2+b^2c^2+c^2a^c)=\dfrac{1}{2}$

Ta có $a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^c)=
1-\dfrac{1}{2}=\dfrac{1}{2}$
 
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