a) Theo BĐT Bunhia:$(\sqrt{x-1}+\sqrt{y-2})^2 \le 2(x+y-3)=2$
=> $\sqrt{x-1}+\sqrt{y-2} \le \sqrt{2}$
Dấu = xảy ra <=> $x=\frac{3}{2},y=\frac{5}{2}$
b)
$(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y)^2 \le (2+3)(2x^2+3y^2) \le 5.5 =25$
=> $2x+3y \le 5$
Dấu = xảy ra <=> $ x = y = 1$