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+x2)/(1+x1x2)](/uploads/image/z/19555111-55-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dlog2%281%2Bx%2F1-x%29%2C%E6%B1%82%E8%AF%81f%28x1%29%2Bf%28x2%29%3Df%5B%28x1%2Bx2%29%2F%281%2Bx1x2%29)
f(x1)+f(x2)=log2[(1+x1)/(1-x1)]+log2[(1+x2)/(1-x2)]
=log2[(1+x1)(1+x2)/(1-x1)(1-x2)]
f[(x1+x2)/(1+x1x2)]
=log2[1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)]
就是要证明(1+x1)(1+x2)/(1-x1)(1-x2)=[1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)
1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)
=(x1x2+x1+x2)/(x1x2-x1-x2)
而(1+x1)(1+x2)/(1-x1)(1-x2)
=(1+x1x2+x1+x2)/(1+x1x2-x1-x2)
显然这个等式不成立,所以题目有错.
=log2[(1+x1)(1+x2)/(1-x1)(1-x2)]
f[(x1+x2)/(1+x1x2)]
=log2[1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)]
就是要证明(1+x1)(1+x2)/(1-x1)(1-x2)=[1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)
1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)
=(x1x2+x1+x2)/(x1x2-x1-x2)
而(1+x1)(1+x2)/(1-x1)(1-x2)
=(1+x1x2+x1+x2)/(1+x1x2-x1-x2)
显然这个等式不成立,所以题目有错.
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