已知c/(a-b)=a/(b-c)=b/(a-c)=k,a,b,c为正数,则函数y=-kx一定过下列哪个点()A.(1,1/2)B.(1,-1/2)C.(1,2)D,(1,-1)

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已知c/(a-b)=a/(b-c)=b/(a-c)=k,a,b,c为正数,则函数y=-kx一定过下列哪个点()A.(1,1/2)B.(1,-1/2)C.(1,2)D,(1,-1)
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已知c/(a-b)=a/(b-c)=b/(a-c)=k,a,b,c为正数,则函数y=-kx一定过下列哪个点()A.(1,1/2)B.(1,-1/2)C.(1,2)D,(1,-1)
已知c/(a-b)=a/(b-c)=b/(a-c)=k,a,b,c为正数,则函数y=-kx一定过下列哪个点()
A.(1,1/2)
B.(1,-1/2)
C.(1,2)
D,(1,-1)

已知c/(a-b)=a/(b-c)=b/(a-c)=k,a,b,c为正数,则函数y=-kx一定过下列哪个点()A.(1,1/2)B.(1,-1/2)C.(1,2)D,(1,-1)
已知c/(a-b)=a/(b-c)=b/(a-c)=k
c=(a-b)k,
a=(b-c)k=[b-(a-b)k]k=bk-ak+bk^2,a(1+k)=bk(1+k)
(a-bk)(k+1)=0,
如果k=-1则y=-kx=x,则四个点都不在其内
所以a=bk,c=bk(k-1)
因为a,b,c为正数,所以k>1
b=(a-c)k=[bk-bk(k-1)]k=bk^2(2-k)
b[1-k^2(2-k)]=0
因为a,b,c为正数,所以k^2(2-k)=1
k^3-2k^2+1=0
(k-1)(k^2-k-1)=0
因为k>1,所以k-1≯0,所以k^2-k-1=0
则k=(1±√5)/2,因为k>1,所以k=(√5+1)/2
所以y=-kx=-(√5+1)x/2,四个点都不在其内
所以题目有问题,如果k=1,
则a=b-c,c=a-b=b-c-b=-c,c=0,但是c=0又与k=1=c/(a-b)矛盾