已知函数F(X)=log2(1+X)/(1-X) 求证f(x1)+f(x2)=f((x1+x2)/(1+x1x2))

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已知函数F(X)=log2(1+X)/(1-X)求证f(x1)+f(x2)=f((x1+x2)/(1+x1x2))已知函数F(X)=log2(1+X)/(1-X)求证f(x1)+f(x2)=f((x1

已知函数F(X)=log2(1+X)/(1-X) 求证f(x1)+f(x2)=f((x1+x2)/(1+x1x2))
已知函数F(X)=log2(1+X)/(1-X) 求证f(x1)+f(x2)=f((x1+x2)/(1+x1x2))

已知函数F(X)=log2(1+X)/(1-X) 求证f(x1)+f(x2)=f((x1+x2)/(1+x1x2))
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))

就把左边右边分别代入..最后肯定相等的咯