Ta có: [tex]x-\sqrt{x+6}=\sqrt{y+6}-y[/tex]
[tex]<=>x+y=\sqrt{x+6}+\sqrt{y+6}[/tex]
[tex]<=>(x+y)^2=x+y+12+2\sqrt{(x+6)(y+6)}[/tex]
Lại có: [tex]2\sqrt{(x+6)(y+6)}\leq x+y+12[/tex] [tex]=>x+y+12+2\sqrt{(x+6)(y+6)}\leq 2(x+y)+24[/tex]
[tex]=>(x+y)^2\leq (x+y)+24[/tex]
Đặt [tex](x+y)^2=t[/tex]
[tex]=>t^2-2t-24\leq 0[/tex]
[tex]=>(t-1)^2-25\leq 0[/tex]
[tex]=>(t-1)^2\leq 25[/tex]
[tex]=>-5\leq t-1\leq 5[/tex]
[tex]=>-4\leq x+y\leq 6[/tex]
Dấu "=" tự chỉ nhé bạn