1. [tex]a+b\geq 2\sqrt{ab}=>ab\leq \frac{(a+b)^2}{4}=1[/tex]
[tex]A=\frac{(a ^2-4).(b^2-4) }{(ab)^2}=\frac{(a-2)(a+2)(b-2)(b+2)}{(ab)^2}=\frac{(a-a-b)(a+2)(b-a-b)(b+2)}{(ab)^2}=\frac{(a+2)(b+2)}{ab}=\frac{ab+2a+2b+4}{ab}=\frac{ab+6}{ab}=1+\frac{8}{ab}\geq 1+\frac{8}{1}=9[/tex]
2. [tex]<=>3\sqrt{x}+7\sqrt{y}=40\sqrt{2}[/tex]
ta suy ra [tex]\sqrt{x}=a\sqrt{2};y=b\sqrt{2}=>3a\sqrt{2}+7b\sqrt{2}=30\sqrt{2}<=>3a+7b=30[/tex]
bộ số duy nhất thỏa mãn là [tex]a=b=3=>x=y=(3\sqrt{2})^2=18[/tex]