Ta có:$x+2y=1⇒x=1-2y$
Khi đó
$A=(1-2y)^2+2y^2$
$=$$1-4y+4y^2+2y^2$
$=$$6y^2-4y+1$
$=$$6(y^2-\dfrac{4}{6}+\dfrac{1}{6})$
$=$$6(y^2-2.\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{18})$
$=$$6(y-\dfrac{1}{3})^2+\dfrac{1}{3}$
Mà $(y-\dfrac{1}{3})^2≥0$
⇒$6(y-\dfrac{1}{3})^2+\dfrac{1}{3}≥\dfrac{1}{3}$
Dấu = xảy ra $⇔y=\dfrac{1}{3}$
Vậy $Min_A=\dfrac{1}{3}$ tại $y=\dfrac{1}{3}$