1) Ta có
$x^2 + y^2 + 2xy + 2x + 2y + 1 = (x+y)^2 + 2(x+y) + 1 = (x+y)^2 + (x+y) + (x+y) + 1 = (x+y)(x+y+1) + (x+y+1) = (x+y+1)^2$
2) Ta có
$(x-y)^2 + 2(y-x) + 1 = (x-y)^2 - 2(x-y) + 1 = (x-y)^2 - (x-y) - (x-y) + 1 = (x-y)(x-y-1) - (x-y-1) = (x-y-1)^2$.