问题 解答题
已知正实数x、y、z、w满足2007x2=2008y2=2009z2=2010w2,且
1
x
+
1
y
+
1
z
+
1
w
=1
,求
2007x+2008y+2009z+2010w
之值.
答案

设2007x2=2008y2=2009z2=2010z2=A,

∴2007x=

A
x
,2008y=
A
y
,2009z=
A
z
,2010w=
A
w

2007
A
=
1
x
2008
A
=
1
y
2009
A
=
1
z
2010
A
=
1
w

2007
A
+
2008
A
+
2009
A
+
2010
A
=
1
x
+
1
y
+
1
z
+
1
w
=1,

A
=
2007
+
2008
+
2009
+
2010

∴2007x+2008y+2009z+2010w=

A
x
+
A
y
+
A
z
+
A
w

=A(

1
x
+
1
y
+
1
z
+
1
w
),

1
x
+
1
y
+
1
z
+
1
w
=1,

∴2007x+2008y+2009z+2010w=A.

2007x+2008y+2009z+2010w
=
A
=
2007
+
2008
+
2009
+
2010

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