Tìm GTNN của [tex]A=x^2+3y^2-xy-3x-3y[/tex]
em cần gấp câu này ạ
@Nữ Thần Mặt Trăng chị giúp em với
$A=x^2+3y^2-xy-3x-3y$
$=(x^2+\dfrac{y^2}4+\dfrac 94-xy-3x+\dfrac{3y}2)+\dfrac{11}4(y^2-\dfrac{18}{11}y+\dfrac{81}{121})-\dfrac{45}{11}$
$=(x-\dfrac y2-\dfrac 32)^2+\dfrac{11}4(y-\dfrac 9{11})^2-\dfrac{45}{11}\ge \dfrac{-45}{11}$
Dấu '=' xảy ra $\Leftrightarrow x=\dfrac{21}{11}; y=\dfrac 9{11}$.
A = [tex]\frac{x}{( x + 2016 )^2}[/tex]
ĐK: $x\ne -2016$.
$A=\dfrac x{(x+2016)^2}=\dfrac{8064x}{8064(x+2016)^2}
\\=\dfrac{(x^2+4032x+2016^2)-(x^2-4032x+2016^2)}{8064(x+2016)^2}
\\=\dfrac{(x+2016)^2-(x-2016)^2}{8064(x+2016)^2}
\\=\dfrac1{8064}-\dfrac{(x-2016)^2}{8064(x+2016)^2}\le \dfrac1{8064}$
Dấu '=' xảy ra $\Leftrightarrow x=2016$ (t/m)
[tex]A=\frac{x}{(x+2016)^{2}}=\frac{1}{(\sqrt{x}+\frac{2016}{\sqrt{x}})^{2}}\leq \frac{1}{(2\sqrt{2016}^{2})}=\frac{1}{8064}[/tex]
dấu = x=2016
Nếu $x\le 0$ thì sao ạ?