Đáp án:
`P=24\sqrt{3}`
Giải thích các bước giải:
$\begin{cases}a^2+ab+\dfrac{b^2}{3}+25(1)\\\dfrac{b^2}{3}+c^2=9 (2)\\c^2+ac+a^2=16(3)\end{cases}$
Cộng từng vế $(2),(3)$ ta được:
$a^2+\dfrac{b^2}{3}+2c^2+ac=25=a^2+ab+\dfrac{b^2}{3}$
$⇔ 2c^2=ab-ac$
$⇔ 2c^2=a(b-c)$
Xét `12.(b^2/3+c^2)(c^2+ac+a^2)`
`=4b^2c^2+12c^4+4ab^2c+12ac^3+4a^2b^2+12a^2c^2`
`=4b^2c^2+6c^2.2c^2+4ab^2c+6ac.2c^2+4a^2b^2+12a^2c^2`
`=4b^2c^2+6c^2.a(b-c)+4ab^2c+6a^2c(b-c)+4a^2b^2+12a^2c^2`
`=4b^2c^2+6abc^2-6ac^3+4ab^2c+6a62bc-6a^2c^2+4a^2b^2+12a^2c^2`
`=(a^2b^2+4b^2c^2+9a^2c^2+4ab^2c+12abc^2+6a^2bc)+(3a^2b^2-3a^2c^2-6abc^2-6ac^3)`
`=(ab+2bc+3ac)^2+3a^2(b^2-c^2)-6ac^2(b+c)`
`=P^2+3a^2(b-c)(b+c)-6ac^2(b+c)`
`=P^2+3a(b+c).2c^2-6ac^2(b+c)`
`=P^2`
Như vậy `P^2=12.9.16=1728`
`⇔ P=24\sqrt{3}` (do P>0)
