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» Gallery » Logical puzzles 011001101001000011
Tři rovnice - difficulty 4.2 (former difficulty 4)
Máme dánu soustavu tří rovnic o třech neznámých x, y, z:
x + y + z = 1
x2 + y2 + z2 = 2
x3 + y3 + z3 = 4
Čemu je roven součin x * y * z ?
Čemu je roven výraz 1/x + 1/y + 1/z ?
x + y + z = 1
x2 + y2 + z2 = 2
x3 + y3 + z3 = 4
Čemu je roven součin x * y * z ?
Čemu je roven výraz 1/x + 1/y + 1/z ?
(x+y+z)3 = (x3+y3+z3) + 6xyz + 3xy2 + 3y2z + 3xz2 + 3yz2 + 3x2z + 3x2y
6xyz = (x+y+z)3 - (x3+y3+z3) - 3*(x2*(y+z) + y2*(x+z) + z2*(x+y)) = (x+y+z)3 - (x3+y3+z3) - 3*((x2+y2+z2)(x+y+z) - (x3+y3+z3)) = 1 - 4 - 3*(2*1 - 4) = 3
xyz = 1/2
(x+y+z)2 = (x2+y2+z2) + 2xy + 2xz + 2yz
(xy + xz + yz) = ((x+y+z)2 - (x2+y2+z2))/2 = (1 - 2)/2 = -1/2
1/x + 1/y + 1/z = (xy + xz + yz)/(xyz) = (-1/2)/(1/2) = -1
6xyz = (x+y+z)3 - (x3+y3+z3) - 3*(x2*(y+z) + y2*(x+z) + z2*(x+y)) = (x+y+z)3 - (x3+y3+z3) - 3*((x2+y2+z2)(x+y+z) - (x3+y3+z3)) = 1 - 4 - 3*(2*1 - 4) = 3
xyz = 1/2
(x+y+z)2 = (x2+y2+z2) + 2xy + 2xz + 2yz
(xy + xz + yz) = ((x+y+z)2 - (x2+y2+z2))/2 = (1 - 2)/2 = -1/2
1/x + 1/y + 1/z = (xy + xz + yz)/(xyz) = (-1/2)/(1/2) = -1
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