好评回答：(1)x=a/(b c)→x 1=(a b c)/(b c)→1/(x 1)=(b c)/(a b c)∴x/(x 1)=[(b c)/(a b c)]·[a/(b c)]=a/(a b c).同理，y/(1 y)=b/(a b c)，z/(1 z)=c/(a b c).三式相加，得x/(1 x) y/(1 y) z/(1 z)=(a b c)/(a b c)=1。(2)x 1/x=3，故x^2/(x^4 x^2 1)=1/[(x^2 1/x^2) 1]=1/[(x 1/x)^2-1]=1/(9-1)=1/8。
初高中数学衔接问题
完整问题：求详细解答
好评回答：(1)x=a/(b c)→x 1=(a b c)/(b c)→1/(x 1)=(b c)/(a b c)∴x/(x 1)=[(b c)/(a b c)]·[a/(b c)]=a/(a b c).同理，y/(1 y)=b/(a b c)，z/(1 z)=c/(a b c).三式相加，得x/(1 x) y/(1 y) z/(1 z)=(a b c)/(a b c)=1。(2)x 1/x=3，故x^2/(x^4 x^2 1)=1/[(x^2 1/x^2) 1]=1/[(x 1/x)^2-1]=1/(9-1)=1/8。