(x+y)(x+y+z)=72
(y+z)(x+y+z)=120
(x+z)(x+y+z)=96
складываем построчно:
(x+y)(x+y+z) + (y+z)(x+y+z) + (x+z)(x+y+z) 72 + 120 + 96
(x+y+z)((x+y)+(y+z)+(x+z)) = 288
(x+y+z)(2x+2y+2z) = 288
(x+y+z)2(x+y+z) = 288
(x+y+z)^2 = 144
x+y+z = +-12
1. x1 = 12-y-z 2. x2 = -12-y-z
подставляем "X1 и "Х2"" в исходную систему:
1. (12-y-z+y)(12-y-z+y+z) = 72 2. (-12-y-z+y)(-12-y-z+y+z) = 72
(y+z)12 = 120 (y+z)(-12) = 120
(12-y-z+z)12 = 96 (-12-y-z+z)(-12) = 96
(12-z)12 = 72 (-12-z)(-12) = 72
(y+z)12 = 120 (y+z)(-12) = 120
(12-y)12 = 96 (12-y)(-12) = 96
12-z = 6 -12-z = -6
y+z = 10 y+z = -10
12-y = 8 -12-y = -8
z = 12-6=6 z = -12+6 = -6
y = 12-8=4 y= -12+8 = -4
x1 = 12-6-4 = 2 x2 = -12-(-6)-(-4)=-2
Ответ: z = 12-6=6 z = -12+6 = -6
y = 12-8=4 y= -12+8 = -4
x1 = 12-6-4 = 2 x2 = -12-(-6)-(-4)=-2