if x+y+z=xyz,prove that:?
830855
What is to be proved? It's not clear.
479850
Firstly, we have to multiply all of the eqautions together.\n\nxy*yz*zx \n\n= 6*9*24 \n\n= (xyz)^2 \n\n= 1296\n\nNext step is to find the square root. (Let V = square root)\n\nV1296\n\n= 36\n\n= xyz
1039077
Let the numbers be x, y and z.\nThen xyz = x + y +z\n.Also the numbers must be small, as this is not possible for large numbers.\nThe other set is -1, -2, -3 and -1,0,1 and 2,0,-2 and \n-n,0,n
1006535
You group all like terms together so it would be \n-5x+5x-3y+y-9z+8z\n-5x+5x that = 0\n-3y+y=-2y\n-9z+8z=-z\nSo you bring it all together and you have\n-z-2y
983933
(x+y+z)(x+y+z) = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 1\n\n(1-x^2 - y^2 - z^2) = 2xy + 2xz + 2yz\n\n(1 - x^2 - y^2 - z^2) \n------------------------ = xy + xz + yz\n.............2\n\nClearly the left hand side is less than 1/2
751073
Oh why you asked same question 2 times ???? \nAny way, the answer is what i explained for you earlier\n\nX = 2∏ + ∏/6 , x = 4∏ + ∏/6 , …., x = 2k∏ + ∏/6 , ( k Є Z) \nX = 2∏ + 5∏/6 , x = 4∏ + 5∏/6 ,….., x = 2k∏ + 5∏/6 , ( k Є Z)\n\n\nGood Luck
835073
okay well\n\n1) combine two to figure out the first variable\nx-2y+z=4\n-x+2y+3z=-4\nso 4z=0; z=0\nthen figure out the next variable with a different set of equations\nlike and multiply so the ys will cancel [take Zs out cause zero]\nx-2y+z=4\n2(2x+y)=6\n\ntherefore\nx-2y=4\n4x+2y=6\n5x=10\nx=2\n\nthen substitute for y\n\n-x+2y=-4\n-2+2y=-4\n2y=-2\ny=-1\n\nso y=-1, x=2, z=0\n\n\ncontinue the process with the rest.
635101
m<Z = 52\nThe three angles add to 180.\nm<X + m<Y + m<Z = 180\nThus,\nm<Y = 180 - 52 - 52\n= 76\nAlso, I assumed you meant that XY is congruent (or equal) to YZ.  Then m<X = m<Z.
677039
(1) { -2x-y-z=0\n(2) { -x-2y-z=0\n(3) { -x-y-2z=0\n\nLet's look at (2):\n-x-2y-z=0 => x=-2y-z\nNow at (3):\n-x-y-2z=0 => x=-y-2z\nNow we can see that\n-2y-z=-y-2z => *** y=z ***\n\nNow let's put these values into (1):\n-2x-y-z=0 => -2x-y-y=0 => -2x-2y=0 => 2x=-2y => *** x=-y ***\n\nFor now, we know that x=-y=-z\nLet's put all of what we know into (2), and express everything using 'x':\n-x-2y-z=0 => -x-(-2x)-(-x)=0 => -x+2x+x=0 => 2x=0 => *** x=0 ***\n\nThus, x=0 => y=-0=0 => z=0\nSo the only solution is (0,0,0).\n\nGood luck!
170974
Prove that for  u,v, and w, p1=s(u)+t(v)+r(w), i.e that you can create a linear combination of vectors such that v=a1v1+a2v2+...+anvn.
567748
you are almost right - or maybe the end of your answer got cut off.\ndz/dx gives\n6xz+3x^2(dz/dx)\n-(yz^3+(3z^2)xy(dz/dx)=0\n\nthen just solve this equation for dz/dx
421674
y = (x mod 5) + 1.\n\nThough that does seem a bit like cheating :)
516632
Let P=A\nwhere P = 2x + 2y and A = xy\nthen\n2x+2y = xy\nMove xy to the left side and 2x to the right side\n2y-xy = -2x\nFactor a y out\ny(2-x) = -2x\nDivide by 2-x\ny = -2x/(2-x)\nThis is positive whenever x is greater than 2.\nAnswer: x = any number greater than 2 and y = -2x/(2-x)
739421
x<1/2(x+y)<y can be proved by splitting it in two parts and proving those:\nx<1/2(x+y) is true because 1/2(x+y) > 1/2(x+x) = x\n1/2(x+y)<y is true because 1/2(x+y) < 1/2(y+y) = y
1172689
Not easy!  let z=x+iy\n acos(z) = -i ln((z±(z²-1)^½)\n\n(err.. I think that's right)\nI'll let you do the others!!
558731
distributive property;\n\na(b+c) = ab + ac
474735
If z = x+iy, then the complex conjugate of z is Z'= x-iy.\nIn your sum the conjugate of 2ix - i = i(2x-1).  The real component is zero and the imaginary component is 2x-1. Then the conjugate of 2ix - i must be i(1-2x).  Similarly, \nin (1-2x) the real component is 1-2x and the imaginary component is zero.  The complex conjugate is 1-2x.