分类：教育/科学-理工学科-数学
问题：数学问题已知mnp=4，m+n+p=3，m平方+n平方+p平方= 
已知mnp=4，m+n+p=3，m平方 + n平方 + p平方=7
求分式1/（mn+3p）+1/(np+3m)+1/(pm+3n)=?
回答：1/（mn+3p）+1/(np+3m)+1/(pm+3n)
=1/(mn+(m+n+p)p)+1/(np+(m+n+p)m)+1/(pm+(m+n+p)n)
=1/(mn+mp+np+p^2)+1/(mn+mp+np+m^2)+1/(mn+mp+np+n^2)
∵(m+n+p)^2=m^2+n^2+p^2+2(mn+mp+np)
∴9=7+2（mn+mp+np)
mn+mp+np=1
1/（mn+3p）+1/(np+3m)+1/(pm+3n)
=1/(1+p^2)+1/(1+m^2)+1/(1+n^2)
=[(1+m^2)(1+n^2)+(1+p^2)(1+n^2)+(1+p^2)(1+m^2)]/[(1+p^2)(1+m^2)(1+n^2)]
=(17+m^2n^2+p^2n^2+p^2m^2)/(24+m^2n^2+p^2n^2+p^2m^2)
∵(mn+pn+pm)^2=m^2n^2+p^2n^2+p^2m^2+2mnp(m+n+p)
∴m^2n^2+p^2n^2+p^2m^2=1-2*4*3=-23
1/（mn+3p）+1/(np+3m)+1/(pm+3n)
=(17-23)/(24-23)
=-6