Chủ đề này có 3 trả lời

[yenhanhtuong]

$\left\{\begin{matrix}
 & x^2+y^2+x^2y^2=1+2xy & \\
 & (x-y)(1+xy)=1-xy &
\end{matrix}\right.$

 

$\left\{\begin{matrix}
 & \sqrt{x+1}+\sqrt{y-1}=4 & \\
 & \sqrt{x+6}+\sqrt{y+4}=6 &
\end{matrix}\right.$​

 

$\left\{\begin{matrix}
 & \sqrt{x^2+y^2}+\sqrt{2xy}=8.\sqrt{2} & \\
 & \sqrt{x}+\sqrt{y}=4 &
\end{matrix}\right.$

[Nguyen Kieu Phuong]

$\left\{\begin{matrix}
 & x^2+y^2+x^2y^2=1+2xy & \\
 & (x-y)(1+xy)=1-xy &
\end{matrix}\right.$

 

 

$\left\{\begin{matrix}
 & \sqrt{x^2+y^2}+\sqrt{2xy}=8.\sqrt{2} & \\
 & \sqrt{x}+\sqrt{y}=4 &
\end{matrix}\right.$

[Issac Newton of Ngoc Tao]

$\left\{\begin{matrix}

 & \sqrt{x+1}+\sqrt{y-1}=4 & \\
 & \sqrt{x+6}+\sqrt{y+4}=6 &
\end{matrix}\right.$​

Đặt $a=\sqrt{x+1};b=\sqrt{y-1}\Rightarrow HPT:\left\{\begin{matrix} a+b=4(1) & \\ \sqrt{a^{2}+5}+\sqrt{b^{2}+5}=6(2) & \end{matrix}\right.;(2)\Rightarrow a^{2}+5=36+b^{2}+5-12\sqrt{b^{2}+5}\Leftrightarrow 3\sqrt{b^{2}+5}=13-2b(do(1))$.      

[yenhanhtuong]

👌👌👌👌👌👌