神马品牌网唐氏综合症临界风险值:数学代数
来源：百度文库 编辑：神马品牌网 时间：2024/05/09 06:43:11
已知a=x/(y+z),b=y/(z+x),c=z/(x+y),x+y+z≠0
求a/(a+1)+b/(b+1)+c/(c+1)
求a/(a+1)+b/(b+1)+c/(c+1)
a/(a+1)+b/(b+1)+c/(c+1)
=x/(y+z)/<x/(y+z)+1>+y/(z+x)/<y/(z+x)+1>+z/(x+y)/<z/(x+y)+1>
=x/(x+y+z)+y/(x+y+z)+z/(x+y+z)
=(x+y+z)/(x+y+z)
=1