1/a+1/b+1/c=1通分：(ab+bc+ca)/abc=1∴ab+bc+ca=abc∴ab+bc+ca-abc=0=ab+bc+ca-a-b-c-abc+a+b+c=ab+bc+ca-a-b-c-abc+1=1-a+(ab+bc+ca-b-c-abc)=1-a+[b(a-1)+c(a-1)-bc(a-1)]=(1-a)(1-b-c+bc)=(1-a)(1-b)(1-c)=0∴a,...