若函数f[(x+1)/x]=(x²+1)/x²+1/x,则函数f(x)的表达式为( )A.f(x)=x²-2x+1 B.f(x)=x²-x+1(x≠0) C f(x)=x²-x+1(x∈R) D f(x)=x²-x+1(x≠1)
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![若函数f[(x+1)/x]=(x²+1)/x²+1/x,则函数f(x)的表达式为( )A.f(x)=x²-2x+1 B.f(x)=x²-x+1(x≠0) C f(x)=x²-x+1(x∈R) D f(x)=x²-x+1(x≠1)](/uploads/image/z/13434658-34-8.jpg?t=%E8%8B%A5%E5%87%BD%E6%95%B0f%5B%28x%2B1%29%2Fx%5D%3D%EF%BC%88x%26%23178%3B%2B1%EF%BC%89%2Fx%26%23178%3B%2B1%2Fx%2C%E5%88%99%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F%E4%B8%BA%EF%BC%88+%EF%BC%89A.f%28x%29%3Dx%26%23178%3B-2x%2B1+B.f%28x%29%3Dx%26%23178%3B-x%2B1%28x%E2%89%A00%29+C+f%28x%29%3Dx%26%23178%3B-x%2B1%28x%E2%88%88R%EF%BC%89+D+f%28x%29%3Dx%26%23178%3B-x%2B1%28x%E2%89%A01%EF%BC%89)
若函数f[(x+1)/x]=(x²+1)/x²+1/x,则函数f(x)的表达式为( )A.f(x)=x²-2x+1 B.f(x)=x²-x+1(x≠0) C f(x)=x²-x+1(x∈R) D f(x)=x²-x+1(x≠1)
若函数f[(x+1)/x]=(x²+1)/x²+1/x,则函数f(x)的表达式为( )
A.f(x)=x²-2x+1 B.f(x)=x²-x+1(x≠0) C f(x)=x²-x+1(x∈R) D f(x)=x²-x+1(x≠1)
若函数f[(x+1)/x]=(x²+1)/x²+1/x,则函数f(x)的表达式为( )A.f(x)=x²-2x+1 B.f(x)=x²-x+1(x≠0) C f(x)=x²-x+1(x∈R) D f(x)=x²-x+1(x≠1)
解答如下
设(x+1)/x=t,则t=1+1/x,f[(x+1)/x]=f(t)=1+1/x²+1/x=1+(t-1)²+(t-1)=t²-t+1,所以函数表达式是
f(x)=x²-x+1因为x是分母,所以x≠0,所以选择B
不明白,再问我.
f[(x+1)/x]
=(x²+1)/x²+1/x
=(1+1/x+1/x^2)
=f(1+1/x)
f(x)=x^2-x+1
且x≠1
选D
令(x+1)/x=t即x=1/(t-1)代入原式得:
f(t)=【(1/(t-1))^2+1】/(1/(t-1))^2+(t-1)
=t^2-t+1
股f(x)的表达式如下:
f(x)=x^2-x+1
我不清楚