作业帮高一函数证明题f(x)=log2 (1+x)/(1-x)(1)求证：f(x1)+f(x2)=f[(x1+x2)/(1+x1x2)](2)若f[(a+b)/(1+ab)]=1,f(-b)=1/2,求f(a)的值
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/09 09:26:42
高一函数证明题f(x)=log2 (1+x)/(1-x)(1)求证：f(x1)+f(x2)=f[(x1+x2)/(1+x1x2)](2)若f[(a+b)/(1+ab)]=1,f(-b)=1/2,求f(a)的值
高一函数证明题
f(x)=log2 (1+x)/(1-x)
(1)求证：f(x1)+f(x2)=f[(x1+x2)/(1+x1x2)]
(2)若f[(a+b)/(1+ab)]=1,f(-b)=1/2,求f(a)的值
高一函数证明题f(x)=log2 (1+x)/(1-x)(1)求证：f(x1)+f(x2)=f[(x1+x2)/(1+x1x2)](2)若f[(a+b)/(1+ab)]=1,f(-b)=1/2,求f(a)的值
(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
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+...
全部展开
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
收起
(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)]