牛B人物进已知:a+b+c=0求证:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0

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牛B人物进已知:a+b+c=0求证:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0牛B人物进已知:a+b+c=0求证:a(1/b+1/c)+b(1/c+1/a)+c(1/a+

牛B人物进已知:a+b+c=0求证:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
牛B人物进
已知:a+b+c=0
求证:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0

牛B人物进已知:a+b+c=0求证:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
证明:由a+b+c=0得a=-(b+c),b=-(a+c),c=-(a+b)
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3
=-(b+c)*(1/b+1/c)-(a+c)*(1/c+1/a)-(a+b)*(1/a+1/b)+3
=(-1-b/c-c/b-1)-(a/c+1+1-c/a)-(1+a/b+b/a+1)+3
=-6-b/c-c/b-a/c-c/a-a/b-b/a+3
=-3-b/c-a/c-c/b-a/b-c/a-b/a
=-3-(b+a)/c-(c+a)/b-(c+b)/a
=-3-(0-c)/c-(0-b)/b-(0-a)/a
=-3+1+1+1
=0

a=-(b+c)
b=-(a+c)
c=-(a+b)
带进去就行了

因a+b+c=0
所以a+b=-c,a+c=-b,b+c=-a
所以a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3
=a/b+a/c+b/c+b/a+c/a+c/b+3
=(a+c)/b+(a+b)/c+(b+c)/a+3
=-b/b-c/c-a/a+3
=-3+3
=0

悲催!不会啊……

a+b=-c a+c=-b b+c=-a
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=a/b+a/c+b/c+b/a+c/a+c/b+3=(b+c)/a+(a+c)/b+(a+b)/c+3= (-a)/a+(-b)/b+(-c)/c=(-1)+(-1)+(-1)+3=0

证明:因为a+b+c=0
所以b+c=0-a=-a a+c=0-b =-b a+b=0-c=-c
化简a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)
=a/b+a/c+b/c+b/a+c/a+c/b
=(b+c)/a+(a+...

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证明:因为a+b+c=0
所以b+c=0-a=-a a+c=0-b =-b a+b=0-c=-c
化简a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)
=a/b+a/c+b/c+b/a+c/a+c/b
=(b+c)/a+(a+c)/b+(a+b)/c
=-1-1-1=-3
-3+3=0
所以a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0

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已知:a+b+c=o推出:a+b=-c a+c=-b b+c=-a
所以:a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
a/b+a/c+b/c+b/a+c/a+c/b=o
(a+c)/b+(a+b)/c+(b+c)/a=-3
结果就出来了

打开括号a/B+A/C+B/C+B/A+C/A+C/B+3
合并同类项 (A+B)/C+(A+C)/B+(B+C)/A+3
根据已知可知A+B=-C A+C=-B B+C=-A
所以原式等于-c/c+-b/b+-c/c+3=-1-1-1+3=0

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3
=(b+c)/a+(a+c)/b+(a+b)/c+3
=(b+c)/a+1+(a+c)/b+1+(a+b)/c+1
=(a+b+c)(1/a+1/b+1/c)
=0

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3
=a(1/b+1/c+1/a)+b(1/b+1/c+1/a)+c(1/b+1/c+1/a)
=(a+b+c)(1/b+1/c+1/a)
=0

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3
=-(b+c)*(1/b+1/c)-(a+c)*(1/c+1/a)-(a+b)*(1/a+1/b)+3
=(-1-b/c-c/b-1)-(a/c+1+1-c/a)-(1+a/b+b/a+1)+3
=-6-b/c-c/b-a/c-c/a-a/b-b/a+3
=...

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a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3
=-(b+c)*(1/b+1/c)-(a+c)*(1/c+1/a)-(a+b)*(1/a+1/b)+3
=(-1-b/c-c/b-1)-(a/c+1+1-c/a)-(1+a/b+b/a+1)+3
=-6-b/c-c/b-a/c-c/a-a/b-b/a+3
=-3-b/c-a/c-c/b-a/b-c/a-b/a
=-3-(b+a)/c-(c+a)/b-(c+b)/a
=-3-(0-c)/c-(0-b)/b-(0-a)/a
=-3+1+1+1
=0

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