Chủ đề này có 1 trả lời

[leduylinh1998]

Giải HPT:

$\left\{\begin{matrix} \sqrt{x+3y}+\sqrt{2x+7y}=10 & \\ (\sqrt{x}+\sqrt{y})\left ( \frac{1}{\sqrt{x+3y}}+\frac{1}{\sqrt{3x+y}} \right )=2 & \end{matrix}\right.$

[chardhdmovies]

Giải HPT:

$\left\{\begin{matrix} \sqrt{x+3y}+\sqrt{2x+7y}=10 & \\ (\sqrt{x}+\sqrt{y})\left ( \frac{1}{\sqrt{x+3y}}+\frac{1}{\sqrt{3x+y}} \right )=2 & \end{matrix}\right.$

có $\sqrt{\frac{x}{x+3y}}=\sqrt{\frac{x}{x+y}.\frac{x+y}{x+3y}}\leq \frac{1}{2}(\frac{x}{x+3y}+\frac{x+y}{x+3y})$

$\sqrt{\frac{y}{x+3y}}=\sqrt{\frac{1}{2}.\frac{2y}{x+3y}}\leq \frac{1}{2}(\frac{1}{2}+\frac{2y}{x+3y})$

$\Rightarrow \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+3y}}\leq \frac{1}{2}(\frac{x}{x+y}+\frac{3}{2})$

tương tự ta có $\frac{\sqrt{x}+\sqrt{y}}{\sqrt{3x+y}}\leq \frac{1}{2}(\frac{y}{x+y}+\frac{3}{2})$

do đó $(\sqrt{x}+\sqrt{y})(\frac{1}{\sqrt{3x+y}}+\frac{1}{\sqrt{x+3y}})\leq 2\Rightarrow x=y$

tới đây dễ rồi

 

NTP