Giải hệ: $\begin{cases}x^3+y^3+7(x+y)xy=8xy\sqrt{2(x^2+y^2) }\\\sqrt{y}-\sqrt{2x-3}=6-2x \end{cases}

V

viethoang1999

Xét PT $(1)$
$\left ( 1 \right )$\Rightarrow $x^3+y^3-xy(x+y)=8xy\left ( \sqrt{2x^2+2y^2}-x-y \right )$
\Leftrightarrow $(x+y)(x-y)^2=8xy.\dfrac{(x-y)^2}{x+y+\sqrt{2x^2+2y^2}}$
\Rightarrow $(x-y)^2\left ( x+y-\dfrac{8xy}{x+y+\sqrt{2x^2+2y^2}} \right )=0$
Với $\left ( x+y-\dfrac{8xy}{x+y+\sqrt{2x^2+2y^2}} \right )=0$
Ta thấy $\dfrac{8xy}{x+y+\sqrt{2x^2+2y^2}} \le \dfrac{8xy}{x+y+x+y}\le \dfrac{2(x+y)^2}{2(x+y)}=x+y$
mà $\dfrac{8xy}{x+y+\sqrt{2x^2+2y^2}}=x+y$
Dấu = xảy ra khi $x=y$
Suy ra $x=y$ thay vào PT $(2)$ \Rightarrow $\sqrt{x}-\sqrt{2x-3}=6-2x$\Rightarrow $x=3$
 
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