题目内容

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.
分析:设2007x2=2008Y2=2009z2=2010z2=A,得到:
A
=
2007
+
2008
+
2009
+
2010
,2007x=
A
x
,2008y=
A
y
,2009z=
A
z
,2010w=
A
w
,将上式代入即可得出答案.
解答:解: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
点评:本题考查了二次根式的化简和求值,先将原式变形,是解此题的关键.
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网