\(A=xy=y\left(1-2y\right)=y-2y^2=-\dfrac{1}{8}\left(16y^2-8y\right)=-\dfrac{1}{8}\left(4y-1\right)^2+\dfrac{1}{8}\le\dfrac{1}{8}\)Dấu bằng xảy ra <=> 4y-1=0 <=> y=\(\dfrac{1}{4}\);x=\(\dfrac{1}{2}\)( vì x+2y=1)
Do đó, A=xy đạt GTLN là \(\dfrac{1}{8}\)<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{4}\end{matrix}\right.\)