第1个回答 2015-10-26
解;
(√n+1-√n )*(√n+1+√n )
=n+1 -n
=1 (平方差公式)
所以 (√n+1-√n )=(√n+1+√n )^-1 (就是倒数)
所以 log(√n+1+√n)^(√n+1-√n )=-1
第2个回答 2015-01-25
f(x)=sin(2x-π/6)+cos^2x=√3/2sin2x-1/2cos2x+1/2cos2x+1/2
=√3/2sin2x+1/2=√3sinxcosx+1/2
f(θ)=1 , sinθcosθ=(1/2) /√3=√3/6
2kπ-π/2<2x<2kπ+π/2, kπ-π/4<x<kπ+π/4递增
第3个回答 推荐于2016-11-26
(1)f(x)=根号3/2sin2x-1/2cos2x+(1+cos2x)/2
=根号3/2sin2x+1/2
f(θ)=根号3/2sin2θ+1/2
=1
sin2θ=-根号3/3
sinθ*cosθ=-根号3/6
(2)2kπ-π/2<=2x<=2kπ+π/2
kπ-π/4<=x<=kπ+π/4
增区间【kπ-π/4,kπ+π/4】 k∈Z本回答被提问者采纳