已知正实数x,y,z,w满足2007x^2=2008y^2=2009z^2=2010w^2,且1/x+1/y+1/z+1/w=1,求（2007x+2008y+2009z+2009w）^1/2的值

Question

已知正实数x,y,z,w满足2007x^2=2008y^2=2009z^2=2010w^2,且1/x+1/y+1/z+1/w=1,求（2007x+2008y+2009z+2009w）^1/2的值

Answer 1

设2007x^2=a,则x=√(a/2007)
同样,y=√(a/2008),z==√(a/2009),w=√(a/2010)
由1/x+1/y+1/z+1/w=1得,
(√2007+ √2008+√2009+√2010)/ √a = 1,即
√a = √2007+ √2008+√2009+√2010
那么
（2007x+2008y+2009z+2009w）^1/2
=(√2007a+ √2008a+√2009a+√2010a)^1/2
=√2007+ √2008+√2009+√2010