1/ Tìm GTLN của M=$ \frac{x\sqrt{y-2011}+ y\sqrt{x-2010}}{xy}$ .Dấu "=" xảy ra khi nào?
  2/ Tìm nghiệm nguyên DƯƠNG của pt: $\begin{cases}a +3b=15 \\ a+b=3^{c} \end{cases}$
1/
Ta có:
$y-2011+2011\ge 2\sqrt{2011}\sqrt{y-2011}\Rightarrow\frac{\sqrt{y-2011}}{y}\le\frac{1}{2\sqrt{2011}}$
$x-2010+2010\ge2\sqrt{2010}\sqrt{x-2010}\Rightarrow\frac{\sqrt{x-2010}}{x}\le\frac{1}{2\sqrt{2010}}$
So: $M=\frac{\sqrt{y-2011}}{y}+\frac{\sqrt{x-2010}}{x}\le \frac{1}{2\sqrt{2011}}+\frac{1}{2\sqrt{2010}}$
$"=" \Leftrightarrow x=4020,y =4022$
2/
Từ pt đầu suy ra: $b=\frac{15-a}{3}$
do a, b nguyên dương nên $a$ chạy từ 0 đến 15 và $\frac{15-a}{3}\in z$
$\Rightarrow b=${$3;6;9;12$}
mặt khác từ pt2 suy ra $15-2b=3^{c}$
thay b lần lượt, suy ra c=2
vậy nghiệm của pt là $(6;3;2)$

1.
$M=\frac{\sqrt{y-2011}}{y}+\frac{\sqrt{x-2010}}{x}\leq \frac{1}{2\sqrt{2011}}+\frac{1}{2\sqrt{2010}}$
Dấu "=" $\Leftrightarrow x=2020; y=4022$

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