đặt [tex]\sqrt{x-1}=a\rightarrow x=a^2+1[/tex] và [tex]\sqrt{y-2}=b\rightarrow y=b^2+2[/tex]
Có:
[tex]A=\frac{\sqrt{x-1}}{x} + \frac{\sqrt{y-2}}{y}=\frac{a}{a^2+1}+\frac{b}{b^2+2}\leq \frac{a}{2a}+\frac{b}{2\sqrt{2}.b}=\frac{1}{2}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}+1}{2\sqrt{2}}[/tex]
Dấu = xr khi x=2 ; y=4