$(1)=>(x-y)^2+3(x+y)^2+\frac{3}{(x+y)^2}=7$$(2):(x+y)+(x-y)+\frac{1}{x+y}=3$$a=x+y ;b=x-y$$=>b^2+3a^2+\frac{3}{a^2}=7(*)$ và $a+b+\frac{1}{a}=3(**)$$(*)=>3(a+\frac{1}{a})^2=13-b^2$thay $(**)$ vào $(*):3(3-b)^2=13-b^2=>b=>a=>x;y$
ĐK:$x+y$ # $0$$(1)=>(x-y)^2+3(x+y)^2+\frac{3}{(x+y)^2}=7$$(2):(x+y)+(x-y)+\frac{1}{x+y}=3$$a=x+y ;b=x-y$$=>b^2+3a^2+\frac{3}{a^2}=7(*)$ và $a+b+\frac{1}{a}=3(**)$$(*)=>3(a+\frac{1}{a})^2=13-b^2$thay $(**)$ vào $(*):3(3-b)^2=13-b^2=>b=>a=>x;y$