Giải hệ pt: $\left\{\begin{matrix} \sqrt{x+\frac{1}{y}}+\sqrt{x+y-3}=3 & \\ 2x+y+\frac{1}{y}=8 & \end{matrix}\right.$
$\left\{\begin{matrix} \sqrt{x+\frac{1}{y}}+\sqrt{x+y-3}=3 & \\ 2x+y+\frac{1}{y}=8 & \end{matrix}\right.$
Started By luuvanthai, 04-01-2019 - 18:08
#2
Posted 04-01-2019 - 20:00
Arthur Pendragon
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đặt $\sqrt{x+\frac{1}{y}}=a$; $\sqrt{x+y-3}=b$ $(a,b \geq0)$. khi đó hệ đã cho tương đương với:
$\left\{\begin{matrix} a+b=3\\ a^2+b^2=5 \end{matrix}\right.$
Edited by Arthur Pendragon, 04-01-2019 - 20:01.
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