Chủ đề này có 1 trả lời

[mango]

Giải HPT sau:

$\left\{\begin{matrix} \sqrt{1+2x_{1}}+\sqrt{1+2x_{2}}+...+\sqrt{1+2x_{2009}}=2009.\sqrt{\frac{2010}{2009}}\\ \sqrt{1-2x_{1}}+\sqrt{1-2x_{2}}+...+\sqrt{1-2x_{2009}}=2009.\sqrt{\frac{2008}{2009}} \end{matrix}\right.$

[ducthinh26032011]

Giải HPT sau:

$\left\{\begin{matrix} \sqrt{1+2x_{1}}+\sqrt{1+2x_{2}}+...+\sqrt{1+2x_{2009}}=2009.\sqrt{\frac{2010}{2009}}\\ \sqrt{1-2x_{1}}+\sqrt{1-2x_{2}}+...+\sqrt{1-2x_{2009}}=2009.\sqrt{\frac{2008}{2009}} \end{matrix}\right.$

Ta đặt:

$A=2\sum_{i=1}^{2009}x_{i}$

Ta có:

$\sum_{i=1}^{2009}(2x_{i}+1+\frac{2010}{2009})\geq 2\sqrt{\frac{2010}{2009}}.\sum_{i=1}^{2009}\sqrt{2x_{1}+1}$

$\Leftrightarrow A\geq 1(1)$

Ta lại có:

$\sum_{i=1}^{2009}(1-2x_{i}+\frac{2008}{2009})\geq 2\sqrt{\frac{2008}{2009}}.\sum_{i=1}^{2009}\sqrt{1-2x_{1}}$

$\Leftrightarrow A\leq 1(2)$

Từ $(1),(2)$ $\Leftrightarrow A=1$

Dấu $"="$ xảy ra khi $x_{i}=\frac{1}{2.2009}(i=\overline{1,2009})$