如题所述
第1个回答 2010-06-03
根号下大于等于0
cosx+2cos²x>=0
cosx(2cosx+1)>=0
cosx<=-1/2,cosx>=0
所以
2kπ+2π/3<=x<=2kπ+4π/3,2kπ-π/2<=x<=2kπ+π/2
所以定义域是
(2kπ-π/2,2kπ+π/2)∪(2kπ+2π/3,2kπ+4π/3)
1111122222/33333
=(1111100000+22222)/(11111*3)
=1111100000/(11111*3)+22222/(11111*3)
=100000/3+2/3
=100002/3
=33334
cosx+2cos²x>=0
cosx(2cosx+1)>=0
cosx<=-1/2,cosx>=0
所以
2kπ+2π/3<=x<=2kπ+4π/3,2kπ-π/2<=x<=2kπ+π/2
所以定义域是
(2kπ-π/2,2kπ+π/2)∪(2kπ+2π/3,2kπ+4π/3)
1111122222/33333
=(1111100000+22222)/(11111*3)
=1111100000/(11111*3)+22222/(11111*3)
=100000/3+2/3
=100002/3
=33334
第2个回答 2010-06-03
201201201÷345345345
=201×1001001÷(345×1001001)
=201÷345
=67×3÷(115×3)
=67/115本回答被提问者采纳
=201×1001001÷(345×1001001)
=201÷345
=67×3÷(115×3)
=67/115本回答被提问者采纳