第24页

信息发布者：
A
$\frac{36}{37}$
$\frac{1}{17}$
$解:∵f(n)+f(\frac{1}{n})=\frac{n^2}{1+n^2}=\dfrac {(\frac {1}{n})^2}{1+(\frac {1}{n})^2}=\frac{n^2}{1+n^2}+\frac{1}{1+n^2}=1，$
$∴原式=\frac{1}{2}+\left[f(2)+f(\frac{1}{2})\right]+\left[f(3)+f(\frac{1}{3})\right]+···+\left[f(n+1)+f(\frac{1}{n+1})\right]=\frac{1}{2}+1×n=\frac{1}{2}+n.$
B
$\frac{1}{x+2023}$
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