Question

Решите систему 1+x+y=xy, 2+y+z=yz, 3+z+x=zx

Answer 1

$\left\{\begin{array}{c}1+x+y=xy,\\2+y+z=yz,\\3+z+x=zx;\end{array}\right. \left\{\begin{array}{c}1+x+y=xy,\\y(z-1)=z+2,\\z(x-1)=x+3;\end{array}\right. \left\{\begin{array}{c}1+x+y=xy,\\y=\frac{z+2}{z-1},\\z=\frac{x+3}{x-1};\end{array}\right. \\ \left\{\begin{array}{c}1+x+y=xy,\\y=\frac{z+2}{z-1},\\z=\frac{x+3}{x-1};\end{array}\right. \left\{\begin{array}{c}1+x+y=xy,\\y=\frac{3x+1}{4},\\z=\frac{x+3}{x-1};\end{array}\right. \left\{\begin{array}{c}1+x+\frac{3x+1}{4}=x\cdot\frac{3x+1}{4},\\y=\frac{3x+1}{4},\\z=\frac{x+3}{x-1};\end{array}\right.$
$4\cdot(1+x)+3x+1=x(3x+1), \\ 7x+5=3x^2+x, \\ 3x^2-6x-5=0, \\ D_1=3^2-3\cdot(-5)=24=(2\sqrt{6})^2, \\ x=\frac{3\pm2\sqrt{6}}{3}, \\ y=\frac{4\pm2\sqrt{6}}{4}, \\ z=\sqrt{6}\pm1.$