lg(√2)^2+1/2lg2*lg5+√((lg√2)^2-lg2+1)

问题描述:

lg(√2)^2+1/2lg2*lg5+√((lg√2)^2-lg2+1)
1/(1+√3)+1/(√3+√5)+1/(√5+√7)+……+1/(√2n-1+√2n+1)化简
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1个回答 分类:数学 2014-11-06

问题解答:

我来补答
lg(√2)^2+1/2lg2*lg5+√((lg√2)^2-lg2+1)
=lg2+1/2lg2*lg5+√((lg√2)^2-2lg√2+1)
=lg2+1/2lg2*lg5+√(lg√2-1)^2
=lg2+1/2lg2*lg5+|lg√2-1|
=lg2+1/2lg2*lg5+1-1/2lg2
=1/2lg2+1/2lg2*lg5+1
=1/2lg2+1/2lg2*lg5+1
=1/2lg2(1+lg5)+1
=1/2lg2(1+1-lg2)+1
=lg2-1/2(lg2)^2+1
=-1/2(lg2-1)^2+3/2
lg2 = 0.3010
所以原式=1.2557
1/(1+√3)+1/(√3+√5)+1/(√5+√7)+……+1/(√2n-1+√2n+1)
=(√3-1)/2+(√3-√5)/2+……+(√2n+1-√2n-1)/2
=1/2[√(2n+1)-1]=[√(2n+1)-1]/2
 
 
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