Cho $a,b>0$ thỏa mãn: $ab(a+2b)=3$. Tìm GTNN của biểu thức: $P=2a+b$
$a^2b+2ab^2-3=0$
$\Rightarrow a=\frac{-b^2+\sqrt{b^4+3b}}{b}=-b+\sqrt{\frac{b^3+3}{b}}$ 
$\Rightarrow P=2\sqrt{\frac{b^3+3}{b}}-b$
Xét $f(b)=F=2\sqrt{\frac{b^3+3}{b}}-b$
Có $f'(b)=\sqrt{\frac{b}{b^3+3}}(2b-\frac{3}{b^2})-1$
$f'(b)=0\Leftrightarrow 2b-\frac 3{b^2}=\sqrt{\frac{b^3+3}{b}}$
$\Leftrightarrow \frac{2b^3-3}{b^2}=\sqrt{\frac{b^3+3}{b}}$
$\Leftrightarrow \frac{4b^6-12b^3+9}{b^3}=b^3+3 \hspace{3cm} (dk:2b^3 \ge 3)$
$\Leftrightarrow b^6-5b^3+3=0\Leftrightarrow b=\sqrt[3]{\frac{5+\sqrt{13}}{2}}$
Lập bảng bt...
Tìm dc $\min F=\sqrt[3]{\frac{3(13\sqrt{13}-35)}2}$ đạt tại $a=\sqrt[3]{\frac{7\sqrt{13}-25}{2}},$ b như trên
bài nhìn đơn giản mà không hề đơn giản chút nào –  tritanngo99 15-07-16 07:45 PM
nh� v?y l� hay r?i –  tritanngo99 15-07-16 12:35 PM
dùng cách cùi bắp :| –  Aerialace 15-07-16 12:24 PM

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