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[dorabesu]

$\left\{\begin{matrix} 1+xy+\sqrt{xy}=x&&\\\frac{1}{x\sqrt{x}}+x\sqrt{y}=\frac{1}{\sqrt{x}}+3\sqrt{y}&&\end{matrix}\right.$

Edited by dorabesu, 26-01-2013 - 22:11.

[19kvh97]

$\left\{\begin{matrix} 1+xy+\sqrt{xy}=x&&\\\frac{1}{x\sqrt{x}}+x\sqrt{y}=\frac{1}{\sqrt{x}}+3\sqrt{y}&&\end{matrix}\right.$

hệ $\Rightarrow \left\{\begin{matrix} xy+\sqrt{xy}+1=x & & \\ 1+x^2\sqrt{xy}=x+3x\sqrt{xy} & & \end{matrix}\right.$
đặt $\sqrt{xy}=a(a\geq 0)$
hệ trở thành$\left\{\begin{matrix} a^2+a+1=x & & \\ 1+x^2a=x+3ax & & \end{matrix}\right.$
$\left\{\begin{matrix} a^2+a+1=x & & \\ 2a^3+3a^2+3=0 & & \end{matrix}\right.$
$\left\{\begin{matrix} a^2+a+1=x & & \\ 2a^3+3a^2+3a=0 & & \end{matrix}\right.$
suy ra $a=0$ suy ra $y=0$ do $x\neq0$ rồi thay vào là ok