(x - y)2 >= 0 since the left hand side is a square. ie x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy
x2 + y2 = x2 - 2xy + y2 + 2xy = (x - y)2 + 2xy = 72 + 2*8 = 49 + 16 = 65 You could, instead, solve the two equations for x and y and substitute, but the above method is simpler.
(2x2+4xy-3)-(x2-2xy-4)Answer: x2+6x+1
(x + y)2 = x2 + 2xy + y2
(x - y - z)(x - y + z)
(x-y)2 is a square so (x-y)2 >= 0 expanding, x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy or 2xy <= x2 + y2
(x - y)2 >= 0 since the left hand side is a square. ie x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy
x2 + y2 = (x + y)2 => x2 + y2 = x2 + 2xy + y2 => 2xy = 0 => xy = 0 So, one of x and y must be 0.
Not sure. There is no particular name for x2 and 2xy (both include x) in x2 + 2xy + y2 + y3
She was a French mathematician. The relationship x4 + 4y4 = (x2 + 2y2 + 2xy)(x2 + 2y2 − 2xy) is named after her.
the difference between 2x2 +4xy-3 and x2-2xy-4 is?
x2 + y2 = x2 - 2xy + y2 + 2xy = (x - y)2 + 2xy = 72 + 2*8 = 49 + 16 = 65 You could, instead, solve the two equations for x and y and substitute, but the above method is simpler.
(2x2+4xy-3)-(x2-2xy-4)Answer: x2+6x+1
(x + y)(x - 3y)
(x + 5y)(x - 3y)
(x + y)2 = x2 + 2xy + y2
(x+y)2 = x2 + 2xy + y2