ab+a+1分之2ax+bc+b+1分之2bx+ca+c+1分之2cx=1
2ax/(ab+a+1)+2bx(bc+b+1)+2cx/(ca+c+1)=1
2x[a/(ab+a+1)+ab/(abc+ab+a)+abc/(abca+abc+ab)]=1
2x[a/(ab+a+1)+ab/(1+ab+a)+1/(a+1+ab)]=1
2x[(a+ab+1)/(ab+a+1)]=1
2x=1
x=1/2.
方程的解为：x=1/2.