0
If:xy = x2 + y2 + 2xyThen:x2 + xy + y2 = 0Do you want to solve it for x?x2 + xy + (y/2)2 = (y/2)2 - y2(x + y/2)2 = y2/4 - y2x + y/2 = ± √(-3y2/4)x = -y/2 ± y√(-3) / 2x = (-y ± yi√3) / 2
(x + y)3 + (x - y)3 = (x3 + 3x2y + 3xy2 + y3) + (x3 - 3x2y + 3xy2 - y3) = 2x3 + 6xy2 = 2x*(x2 + 3y2)
3 x y x y = 3y2
y = x2 + 3
3x2+6xy+3y2= 3(x2+2xy+y2)= 3(x+y)2
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The GCF is 3.
y=x2+3 x1=1 x2=2 y(x1) = 1*1+3 = 4 y(x2) = 2*2+3 = 7 x2/x1 = 2, While y2/y1 = 7/4 !=2, and thus the function is nonlinear.
If:xy = x2 + y2 + 2xyThen:x2 + xy + y2 = 0Do you want to solve it for x?x2 + xy + (y/2)2 = (y/2)2 - y2(x + y/2)2 = y2/4 - y2x + y/2 = ± √(-3y2/4)x = -y/2 ± y√(-3) / 2x = (-y ± yi√3) / 2
(x + y)3 + (x - y)3 = (x3 + 3x2y + 3xy2 + y3) + (x3 - 3x2y + 3xy2 - y3) = 2x3 + 6xy2 = 2x*(x2 + 3y2)
3y2
8y^3 - 3y^2 = y^2(8y - 3)
Yes. If x is not divisible by 3 then x leaves a remainder of 1 or 2 when it is divided by 3. That is, x is of the form 3y+z where z = 1 or 2. Then x2 = (3y+z)2 = 9y2 + 6yz + z2 = 3(3y2 + 2yz) + z2 The first part of this expression is clearly a multiple of 3, but z2 is not. Whether z = 1 or 2, z2 leaves a remainder of 1 when divided by 3.
f(x) = x2 + 3 ----> f(5) = (5)2 + 3 ----> f(5) = 28
Since 3y2 can evenly divide 24y3, 3y2 must be the greatest common factor, since the greatest common factor cannot be larger than the smaller of the two numbers. But, we can also look at a factorization of both numbers. We can tell that 3y2 = 3 x y2. We can tell that 24y3 = 3 x 8 x y3 = 3 x 23 x y2 x y. These numbers have in common 3 and y2, so 3y2 is the greatest common factor.
3 x y x y = 3y2