Đáp án:
Giải thích các bước giải:
Từ $2c^2=a(b-c)⇔ab=2c^2+ac$
$(ab+2bc+3ac)^2-4(b^2+3c^2)(c^2+ac+a^2)$
$=(a^2b^2+4b^2c^2+9a^2c^2+4ab^2c+12abc^2+6a^2bc)-(4b^2c^2+4ab^2c+4a^2b^2+12c^4+12ac^3+12a^2c^2)$
$=-3a^2b^2-3a^2c^2+12abc^2+6a^2bc-12c^4-12ac^3$
$=-3(2c^2+ac)^2-3a^2c^2+12(2c^2+ac)c^2+6(2c^2+ac).ac-12c^4-12ac^3$
$=-3(4c^4+4ac^3+a^2c^2)-3a^2c^2+24c^4+12ac^3+12ac^3+6a^2c^2-12c^4-12ac^3$
$=-12c^4-12ac^3-3a^2c^2+3a^2c^2+12c^4+12ac^3$
$=0(đpcm)$
