连词成句，第四个小题和第六个小题怎么做？谢谢_百度知道
连词成句，第四个小题和第六个小题怎么做？谢谢
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提问者采纳
I want some milk.
We have PE in the morning.
提问者评价
你的回答完美的解决了我的问题，谢谢！
来自：作业帮
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I want some milk.We have PE in the morning.
4. I want some milk. 6. We have PE in the morning
I want some milk
We have PE in the morning
lwant some milk
I want some milk.
We have PE in the morning.
4.2、1、4、3
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出门在外也不愁尤其第4个怎么做呢,_百度作业帮
尤其第4个怎么做呢,
&尤其第4个怎么做呢,
①：当x1=x2=0时,有[f(0)]^2+[g(0)]^2=g(0),即+[g(0)]^2=g(0),所以,g(0)=0或1,假设,g(0)=0,则：当x1=x2=1时,有[f(1)]^2+[g(1)]^2=g(0),即1+[g(1)]^2=0,不可能,所以,只能g(0)=1成立.②：当x1=x2=-1时,有f(-1)*f(-1)+g(-1)*g(-1)=g(0),即g(-1)=0;当x1=1,x2=-1时,有f(1)*f(-1)+g(1)g(-1)=g(2),即g(2)=0.所以,g(2)不能等于2.③：因为上述可有[f(x)]^2+[g(x)]^2=g(0)=1,所以,[f(x)]^2+[g(x)]^2=1成立.④：因为,[f(x)]^2+[g(x)]^2=1,所以,[f(x)]^2≤1；[g(x)]^2≤1,(当自然数n>2时) 所以,|[f(x)]^n|≤[f(x)]^2；|[g(x)]^n|≤[g(x)]^2,所以,|[f(x)]^n|+|[g(x)]^n|≤[f(x)]^2+[g(x)]^2=1,即有：[f(x)]^n+[g(x)]^n≤|[f(x)]^n|+|[g(x)]^n|≤1,当x=0时,[f(0)]^n+|[g(0)]^n=1,所以,[f(0)]^n+|[g(0)]^n有最大值1.综上,选D.
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