Question

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

Answer 1

$left{begin{array}{c}1+x+y=xy,\2+y+z=yz,\3+z+x=zx;end{array}right. left{begin{array}{c}1+x+y=xy,\y(z-1)=z+2,\z(x-1)=x+3;end{array}right. left{begin{array}{c}1+x+y=xy,\y=frac{z+2}{z-1},\z=frac{x+3}{x-1};end{array}right. \ left{begin{array}{c}1+x+y=xy,\y=frac{z+2}{z-1},\z=frac{x+3}{x-1};end{array}right. left{begin{array}{c}1+x+y=xy,\y=frac{3x+1}{4},\z=frac{x+3}{x-1};end{array}right. left{begin{array}{c}1+x+frac{3x+1}{4}=xcdotfrac{3x+1}{4},\y=frac{3x+1}{4},\z=frac{x+3}{x-1};end{array}right.$
$4cdot(1+x)+3x+1=x(3x+1), \ 7x+5=3x^2+x, \ 3x^2-6x-5=0, \ D_1=3^2-3cdot(-5)=24=(2sqrt{6})^2, \ x=frac{3pm2sqrt{6}}{3}, \ y=frac{4pm2sqrt{6}}{4}, \ z=sqrt{6}pm1.$