[tex]\left\{\begin{matrix} 6x+\frac{3}{x+y}=13\\ 12(x^2+xy+y^2)+\frac{9}{(x+y)^2}=85 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} 3(x-y)+3(x+y+\frac{1}{x+y})=13\\ 9[(x+y)^2+\frac{1}{(x+y)^2}]+3(x-y)^2=85 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} 3(x-y)+3(x+y+\frac{1}{x+y})=13\\ 9(x+y+\frac{1}{x+y})^2+3(x-y)^2=103 \end{matrix}\right.[/tex]
Đặt [tex]a=x-y,b=x+y+\frac{1}{x+y}[/tex]. Hệ trở thành [TEX]\left\{\begin{matrix} 3a+3b=13\\ 3a^2+9b^2=103 \end{matrix}\right.[/TEX]
Hệ này thì dễ rồi nhé.