Đáp án + Giải thích các bước giải:
#$hyn$
$a+b+c=0$
$⇔(a+b+c)^2=0$
$⇔a^2+b^2+c^2+2ab+2ac+2bc=0$
Ta có: $a^2+b^2+c^2=2$
$⇔2ab+2ac+2bc=-2$
$⇔ab+ac+bc=-1$
$⇔(ab+ac+bc)^2=1$
$⇔a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2=1$
$⇔a^2b^2+a^2c^2+b^2c^2+2ab.(a+b+c)=1$
Ta có: $a+b+c=0$
$⇔a^2b^2+a^2c^2+b^2c^2+0=1$
$⇔2a^2b^2+2a^2c^2+2b^2c^2=2$
Ta lại có: $a^2+b^2+c^2=2$
$⇔(a^2+b^2+c^2)^2=4$
$⇔a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=4$
Ta có: $2a^2b^2+2a^2c^2+2b^2c^2=2$
$⇔a^4+b^4+c^4=2$
