高一函数证明题f(x)=log2 (1+x)/(1-x)(1)求证：f(x1)+f(x2)=f[(x

答案:2 悬赏:80 手机版

解决时间 2021-03-02 16:20

提问者网友：温柔港
2021-03-02 10:29

最佳答案

五星知识达人网友：十年萤火照君眠
2021-03-02 11:53

(1)f(x1)+f(x2) =log2(1+x1)/(1-x1)+log2(1+x2)/(1-x2) =log2[(x1+1)(x2+1)/(x1-1)(x2-1)] 若x=(x1+x2)/(1+x1x2) 则(1+x)/(1-x) =[1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)] 上下乘(1+x1x2) =(1+x1x2+x1+x2)/(1+x1x2-x1-x2) =(x1+1)(x2+1)/(x1-1)(x2-1) 所以f((x1+x2)/(1+x1x2))=log2[(x1+1)(x2+1)/(x1-1)(x2-1)] 所以f(x1)+f(x2)=f((x1+x2)/(1+x1x2)) (2)由f(x)=log2 (1+x/1-x),x属于（-1,1）则f(-x)=log2 (1-x/1+x)=log2（1+x/1-x）^(-1)=-log2 1+x/1-x=-f(x) 又x属于（-1,1）,定义域关于原点对称则f(x)是奇函数则f(b)=-f(-b)=-1/2又f[(a+b)/(1+ab)]=f(a)+f(b)=f(a)-1/2=1则f(a)=3/2======以下答案可供参考======供参考答案1:1）左边=f(x1)+f(x2)=log2[(x1+1)/(1-x1)]+log2[(x2+1)/(1-x2)] =log2{[(x1+1)(x2+1)]/[(1-x1)(1-x2)]} =log2[(x1x1+x1+x2+1)/(x1x2+1-x1-x2)] 右边=f[(x1+x2)/(1+x1x2)]=log2[(1+x1x2+x1+x2)/(1+x1x2)] / [(1+x1x2-x1-x2)/(1+x1x2)] =log2[(x1x1+x1+x2+1)/(x1x2+1-x1-x2)]=左边供参考答案2:(1)f(x1)+f(x2)=log2[(1+x1)/(1-x1)]+log2[(1+x2)/(1-x2)] =log2[(1+x1+x2+x1x2)/(1-x1-x2+x1x2)] f[(x1+x2)/(1+x1x2)]=log2{[1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)]} =log2{[(1+x1x2+x1+x2)/(1+x1x2)]/[(1+x1x2-x1-x2)/(1+x1x2)]} =log2[(1+x1+x2+x1x2)/(1-x1-x2+x1x2)] 所以,f(x1)+f(x2)=f[(x1+x2)/(1+x1x2)]

全部回答

1楼网友：几近狂妄
2021-03-02 12:34

和我的回答一样，看来我也对了

我要举报

如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！