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已知x/(y+z+u)=y/(z+u+x)=z/(u+x+y)=u/(x+y+z),求(x+y)/(z+u)+(y+z)/(x+u)+(z+u)/(x+y)+(u+x)/(y+z)
题目详情
已知x/(y+z+u)=y/(z+u+x)=z/(u+x+y)=u/(x+y+z),求(x+y)/(z+u)+(y+z)/(x+u)+(z+u)/(x+y)+(u+x)/(y+z)
▼优质解答
答案和解析
由合分比定理得
x/(y+z+u)=y/(z+u+x)=z/(u+x+y)=u/(x+y+z)=(x+y+z+u)/(3x+3y+3z+3u)=1/3
∴3x=y+z+u.(1)
3y=z+u+x.(2)
3z=u+x+y.(3)
3u=x+y+z.(4)
(1)-(2),得
3(x-y)=y-x
∴x=y
同理y=z,z=u
∴x=y=z=u
代入
(x+y)/(z+u)+(y+z)/(x+u)+(z+u)/(x+y)+(u+x)/(y+z)
=1+1+1+1
=4
x/(y+z+u)=y/(z+u+x)=z/(u+x+y)=u/(x+y+z)=(x+y+z+u)/(3x+3y+3z+3u)=1/3
∴3x=y+z+u.(1)
3y=z+u+x.(2)
3z=u+x+y.(3)
3u=x+y+z.(4)
(1)-(2),得
3(x-y)=y-x
∴x=y
同理y=z,z=u
∴x=y=z=u
代入
(x+y)/(z+u)+(y+z)/(x+u)+(z+u)/(x+y)+(u+x)/(y+z)
=1+1+1+1
=4
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