作业帮已知[(a-b)(b-c)(c-a)]/[(a+b)(b+c)(c+a)]=5/132,求a/(a+b)+b/(b+c)+c/(c+a)的值
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![已知[(a-b)(b-c)(c-a)]/[(a+b)(b+c)(c+a)]=5/132,求a/(a+b)+b/(b+c)+c/(c+a)的值](/uploads/image/z/10361162-2-2.jpg?t=%E5%B7%B2%E7%9F%A5%5B%28a-b%29%28b-c%29%28c-a%29%5D%2F%5B%28a%2Bb%29%28b%2Bc%29%28c%2Ba%29%5D%3D5%2F132%2C%E6%B1%82a%2F%28a%2Bb%29%2Bb%2F%28b%2Bc%29%2Bc%2F%28c%2Ba%29%E7%9A%84%E5%80%BC)
已知[(a-b)(b-c)(c-a)]/[(a+b)(b+c)(c+a)]=5/132,求a/(a+b)+b/(b+c)+c/(c+a)的值
已知[(a-b)(b-c)(c-a)]/[(a+b)(b+c)(c+a)]=5/132,求a/(a+b)+b/(b+c)+c/(c+a)的值
已知[(a-b)(b-c)(c-a)]/[(a+b)(b+c)(c+a)]=5/132,求a/(a+b)+b/(b+c)+c/(c+a)的值
由已知变形,得
(a-b)/(a+b)+(b-c)/(b+c)+(c-a)/(c+a)=5/132
-(a-b)/(a+b)-(b-c)/(b+c)-(c-a)/(c+a)=-5/132
(b-a)/(a+b)+(c-b)/(b+c)+(a-c)/(c+a)=-5/132
(b+a-2a)/(a+b)+(c+b-2b)/(b+c)+(a+c-2c)/(c+a)=-5/132
1-2a/(a+b)+1-2b/(b+c)+1-2c/(c+a)=-5/132
-2a/(a+b)-2b/(b+c)-2c/(c+a)=-5/132-3=-401/132
所以：a/(a+b)+b/(b+c)+c/(c+a)=401/264.
另一解法
取a=0
则得到(b-c)/(b+c)=5/132=k
而b/(b+c)=(k+1)/2
所求式m=1+b/(b+c)=(k+3)/2
猜想上式成立
2m-3=a/(a+b)+b/(b+c)+c/(a+c)-b/(a+b)-c/(b+c)-a/(a+c)=(a-b)/(a+b)+(b-c)/(b+c)+(c-a)/(a+c)=k
令x=(a-b)/(a+b) y=(b-c)/(b+c) z=(c-a)/(c+a)
则2m-3=x+y+z=k=xyz
实际上 x+y+z=[(a-b)(b+c)(a+c)+(b-c)(a+c)(a+b)+(c-a)(a+b)(b+c)]/(a+b)(b+c)(c+a)
分母=(a+c)((a-b)(b+c)+(b-c)(a+b))+(c-a)(a+b)(b+c)=2b(a+c)(a-c)+(c-a)(a+b)(b+c)=(a-b)(b-c)(c-a)
所以得证
即所求=(5/132+3)/2=401/264