toán cực trị

C

congchuaanhsang

ĐKXĐ x \geq 2011

Áp dụng Cauchy ta có:

M = $\dfrac{\sqrt{2011(x-2010)}}{\sqrt{2011}(x+1)}$ + $\dfrac{\sqrt{2010(x-2011)}}{\sqrt{2010}(x-1)}$

\leq $\dfrac{x+1}{2\sqrt{2011}(x+1)}+\dfrac{x-1}{2\sqrt{2010}(x-1)}$

\Leftrightarrow M \leq $\dfrac{1}{2\sqrt{2011}}+\dfrac{1}{2\sqrt{2010}}$

$M_{max}=\dfrac{1}{2\sqrt{2011}}+\dfrac{1}{2\sqrt{2010}}$ \Leftrightarrow $x=4021$ (t/m ĐKXĐ)
 
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