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> 设p=a-ba+b,q=b-cb+c,r=c-ac+a,其中a+b,b+c,c+a全不为零.证明:(1+p)(1+q)(1+r)=(1-p)(1-q)(1-r).-数学
设p=a-ba+b,q=b-cb+c,r=c-ac+a,其中a+b,b+c,c+a全不为零.证明:(1+p)(1+q)(1+r)=(1-p)(1-q)(1-r).-数学
题目简介
设p=a-ba+b,q=b-cb+c,r=c-ac+a,其中a+b,b+c,c+a全不为零.证明:(1+p)(1+q)(1+r)=(1-p)(1-q)(1-r).-数学
题目详情
设
p=
a-b
a+b
,
q=
b-c
b+c
,
r=
c-a
c+a
,其中a+b,b+c,c+a全不为零.证明:(1+p)(1+q)(1+r)=(1-p)(1-q)(1-r).
题型:解答题
难度:中档
来源:不详
答案
证明:1+p=1+
class="stub"a-b
a+b
=
class="stub"2a
a+b
,
1-p=1-
class="stub"a-b
a+b
=
class="stub"2b
a+b
,
同理1+q=
class="stub"2b
c+b
,1-q=
class="stub"2c
c+b
,
1+r=
class="stub"2c
c+a
,1-r=
class="stub"2a
c+a
,
∴
class="stub"左
右
=
(1+q)(1+p)(1+r)
(1-p)1-q)(1-r)
=
class="stub"2a
a+b
+
class="stub"2b
c+b
+
class="stub"2c
c+a
class="stub"2b
a+b
+
class="stub"2c
c+b
+
class="stub"2a
c+a
=1,
∴(1+p)(1+q)(1+r)=(1-p)(1-q)(1-r).
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题目简介
设p=a-ba+b,q=b-cb+c,r=c-ac+a,其中a+b,b+c,c+a全不为零.证明:(1+p)(1+q)(1+r)=(1-p)(1-q)(1-r).-数学
题目详情
答案
1-p=1-
同理1+q=
1+r=
∴
∴(1+p)(1+q)(1+r)=(1-p)(1-q)(1-r).