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> 已知a+ba-b=b+c2(b-c)=c+a3(c-a),a,b,c互不相等.求证:8a+9b+5c=0.-数学
已知a+ba-b=b+c2(b-c)=c+a3(c-a),a,b,c互不相等.求证:8a+9b+5c=0.-数学
题目简介
已知a+ba-b=b+c2(b-c)=c+a3(c-a),a,b,c互不相等.求证:8a+9b+5c=0.-数学
题目详情
已知
a+b
a-b
=
b+c
2(b-c)
=
c+a
3(c-a)
,a,b,c互不相等.求证:8a+9b+5c=0.
题型:解答题
难度:中档
来源:不详
答案
证明:设
class="stub"a+b
a-b
=
class="stub"b+c
2(b-c)
=
class="stub"c+a
3(c-a)
=k,则
a+b=k(a-b),b+c=2k(b-c),
(c+a)=3k(c-a).
所以6(a+b)=6k(a-b),
3(b+c)=6k(b-c),
2(c+a)=6k(c-a).以上三式相加,得
6(a+b)+3(b+c)+2(c+a)
=6k(a-b+b-c+c-a),
即8a+9b+5c=0.
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已知y=a-a2x,z=a-a2y,求证:x=a-a2
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若1a+1b=1c,则a2+b2+c2=(a+b-c)2.-
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题目简介
已知a+ba-b=b+c2(b-c)=c+a3(c-a),a,b,c互不相等.求证:8a+9b+5c=0.-数学
题目详情
答案
a+b=k(a-b),b+c=2k(b-c),
(c+a)=3k(c-a).
所以6(a+b)=6k(a-b),
3(b+c)=6k(b-c),
2(c+a)=6k(c-a).以上三式相加,得
6(a+b)+3(b+c)+2(c+a)
=6k(a-b+b-c+c-a),
即8a+9b+5c=0.