cho 2 số x,y tm $\begin{cases}x>0>y \\ \frac{x^{2}}{2y}-3x+6y-\frac{4y^{2}}{x}-4\leq \frac{6}{xy} \end{cases}$
tìm $Min$
P=$2x^{4}+32y^{4}+4x^{2}y^{2}-2x^{2}-8y^{2}+\frac{1}{x^{2}}+\frac{1}{4y^{2}}-5$
Hình như điều kiện không hợp lí –  Thanh Long 24-05-16 05:46 PM
vậy a viết ra r chụp ảnh -_- –  Confusion 16-05-16 11:13 PM
nại gõ nắm –  Ruanyu Jian 15-05-16 10:25 AM
a lm lun đi ạ –  Nguyễn Nhung 15-05-16 09:37 AM
(x;y) = (1;-12) –  Ruanyu Jian 15-05-16 09:28 AM
Min = -2 –  Ruanyu Jian 15-05-16 09:28 AM
xong xét tính liên tục và đồng biến –  Ruanyu Jian 15-05-16 09:27 AM
em đặt ẩn, phân tích ra –  Ruanyu Jian 15-05-16 09:27 AM
Cách này hơi dài.ai có cách khác up lên cho mk nha...
gt$\Rightarrow x^{3}-6x^{2}y+12xy^{2}-8y^{3}-8xy\geq12\Leftrightarrow (x-2y)^{3}-8xy\geq12$(1)
Đặt $2y=-a,a>0$
(1)tt:$12\leq (x+a)^{3}+4xa\leq (x+a)^{3}+(x+a)^{2}\Rightarrow x+a\geq2$
Ta có:$P=2(x^{4}+a^{4})+x^{2}a^{2}-2(x^{2}+a^{2})+(\frac{1}{x^{2}}+\frac{1}{a^{2}})-5$
$=2(x^{2}+a^{2})^{2}-2(x^{2}+a^{2})-3x^{2}a^{2}+(\frac{1}{x^{2}}+\frac{1}{a^{2}})-5$
$\geq 2(x^{2}+a^{2})^{2}-2(x^{2}+a^{2})-\frac{3}{4}(x^{2}+a^{2})^{2}+\frac{4}{x^{2}+a^{2}}-5$
$=\frac{5}{4}(x^{2}+a^{2})^{2}-2(x^{2}+a^{2})+\frac{4}{x^{2}+a^{2}}$
Đặt $t=x^{2}+a^{2},t\geq2$
$\Rightarrow P\geq \frac{5}{4}t^{2}-2t+\frac{4}{t}-5$
Ta sẽ cm $Min P=-2\Leftrightarrow \frac{5}{4}t^{2}-2t+\frac{4}{t}-5\geq-2$
$\Leftrightarrow(t-2)(5t^{2}+2t-8)\geq0$(luôn đúng với mọi $t\geq2$)
Dấu''='' xra$\Leftrightarrow t=2\Leftrightarrow (x;y)=(1;\frac{-1}{2})$
mn vote dùm mk nha... –  Bloody's Rose 25-05-16 11:46 AM

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