计算x/(y+z)=y/(x+y+1)=z/(x+y-1)的值同上.说得清楚详细些,千万千万不要跳步骤,反正要让我看得懂.答案上是1/2和-1好象要分类讨论的,什么当x+y+z=0时,x+y+z≠0时,

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/14 22:28:56
计算x/(y+z)=y/(x+y+1)=z/(x+y-1)的值同上.说得清楚详细些,千万千万不要跳步骤,反正要让我看得懂.答案上是1/2和-1好象要分类讨论的,什么当x+y+z=0时,x+y+z≠0时,
xRN@4P*)&eUy) jDF-Q B +~111Lsܙ|@8=UJj kRRMi x$f2"G-62t6=ְ= QzxBM'GO4:H9wtɨ2=`br%f 0£-uVl߂OJ=,o8駎877o]JB&Ǝ}<2UYDB(F&;dƘ9s"6u Ef/ةz"]Wh4[-[v~,d3ԲQ&e6zA \3utyUA|@.%N͎9+k}|Y%t

计算x/(y+z)=y/(x+y+1)=z/(x+y-1)的值同上.说得清楚详细些,千万千万不要跳步骤,反正要让我看得懂.答案上是1/2和-1好象要分类讨论的,什么当x+y+z=0时,x+y+z≠0时,
计算x/(y+z)=y/(x+y+1)=z/(x+y-1)的值
同上.
说得清楚详细些,千万千万不要跳步骤,反正要让我看得懂.
答案上是1/2和-1
好象要分类讨论的,什么当x+y+z=0时,x+y+z≠0时,

计算x/(y+z)=y/(x+y+1)=z/(x+y-1)的值同上.说得清楚详细些,千万千万不要跳步骤,反正要让我看得懂.答案上是1/2和-1好象要分类讨论的,什么当x+y+z=0时,x+y+z≠0时,
设x/(y+z)=y/(x+y+1)=z/(x+y-1)=k
则有x=k(y+z)
y=k(x+y+1)
z=k(x+y-1)
三式相加得(x+y+z)=2k(x+y+z)
所以 k=1/2
也就是x/(y+z)=y/(x+y+1)=z/(x+y-1)=k=1/2
我考虑到这一点了,不好意思没写.补充如下:当x+y+z=0时,y+z=-x
x/(y+z)=y/(x+y+1)=z/(x+y-1)=x/(y+z)
=x/(-x)=-1
综上得到x/(y+z)=y/(x+y+1)=z/(x+y-1)=1/2或-1