(x+y)2 = x2 + 2xy + y2
If th equestion meant: (x+y+z)^2The expansion is:(x+y+z)^2= x^2+2xy+y^2+2yz+z^2+2zx
(x-y)2 is a square so (x-y)2 >= 0 expanding, x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy or 2xy <= x2 + y2
Assuming y' is dy/dx, y = x^4/4 + yx^2
4
Only when X or Y = 0 That is because (x + y) squared = x^2 + 2xy + y^2 x^2 + 2xy + y^2 = x^2 + y^2 when x or y = 0
(x+y)2 = x2 + 2xy + y2
2(x - 1)(2x + y)
3x2+6xy+3y2= 3(x2+2xy+y2)= 3(x+y)2
That would be very hard to answer, because it's not. (x + y)2 = x2 + y2 + 2xy
You can expand (x + y)2 into x2 + 2xy + y2
(x - y)2 = x2 - 2xy + y2
If th equestion meant: (x+y+z)^2The expansion is:(x+y+z)^2= x^2+2xy+y^2+2yz+z^2+2zx
2 (2x2 + 2x + xy + y ) it's 2 (2x + y)(x + 1) if you're doing A+
Area of a square = side length squared (x+y)2 (x+y)(x+y) x2+xy+yx+y2 Area = x2+2xy+y2
(3x - y)(x + y)(3x - y)(x + y)
(x-y)2 is a square so (x-y)2 >= 0 expanding, x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy or 2xy <= x2 + y2