1、巧算:(1-1/2^2)×(1-1/3^2)×(1-1/4^2)×……×(1-1/2008^2)

问题描述:

1、巧算:(1-1/2^2)×(1-1/3^2)×(1-1/4^2)×……×(1-1/2008^2)
2、已知2009-a的绝对值与a-2010的算术平方根的和为a,求a-2009^2+15的算术平方根.

(1)
(1-1/2^2)(1-1/3^2)(1-1/4^2)...(1-1/2008^2)
=(2^2-1)/2^2 ×(3^2-1)/3^2 × (4^2-1)/4^2...(2008^2-1)/2008^2
=[(2+1)(2-1)(3+1)(4-1)(4+1)(4-1)...(2008+1)(2008-1)]/(2^2 × 3^2 × 4^2...2008^2)
=(3×1×4×2×5×3×6×4×7×5...2009×2007)/(2^2 × 3^2 × 4^2...2008^2)
=(3×4×5×6...2009×1×2×3...2007)/(2^2 × 3^2 × 4^2...2008^2)
=(1×2×3^2×4^2×5^2×...2007^2×2008×2009)/(2^2 × 3^2 × 4^2...2008^2)
=(1×2×2008×2009)/(2^2×2008^2)=2009/(2×2008)=2009/5016
(2)
因为√(a-2010)>=0
a>=2010
|2009-a|+√(a-2010)=a
a-2009+√(a-2010)=a
√(a-2010)=2009
a-2010=2009^2
a-2009^2=2010
√(a-2009^2+15)
=√(2010+15)
=√2025
=45