Решите систему уравнения (x+y)*(x+y+z)=72 (y+z)*(x+y+z)=120 (z+x)*(x+y+z)=96
101
289
Ответы на вопрос:
Нужно сложить каждую строчку (x+y)(x+y+z) + (y+z)(x+y+z) + (x+z)(x+y+z) 72 + 120 + 96 (x+y++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 x1 = 12-y-z x2 = -12-y-z далее подставляем 2-y-z+y)(12-y-z+y+z) = 72 (-12-y-z+-y-z+y+z) = 72 (y+z)12 = 120 (y+) = 120 (12-y-z+z)12 = 96 (-12-y-z+) = 96 (12-z)12 = 72 (-12-) = 72 (y+z)12 = 120 (y+) = 120 (12-y)12 = 96 (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 = -)=-2 z = 12-6=6 z 1= -12+6 = -6 y = 12-8=4 y1= -12+8 = -4 x1 = 12-6-4 = 2 x2 = -)=-2
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