已知函数f(x)=x²/1+x² ,求f(2)+f(1/2 )+f(3)+f(1/3 )+……+f(2012)+f(1/2012 )的值
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![已知函数f(x)=x²/1+x² ,求f(2)+f(1/2 )+f(3)+f(1/3 )+……+f(2012)+f(1/2012 )的值](/uploads/image/z/5901469-61-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dx%26%23178%3B%2F1%2Bx%26%23178%3B+%2C%E6%B1%82f%282%29%2Bf%281%2F2+%29%2Bf%283%29%2Bf%281%2F3+%29%2B%E2%80%A6%E2%80%A6%2Bf%282012%29%2Bf%281%2F2012+%29%E7%9A%84%E5%80%BC)
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已知函数f(x)=x²/1+x² ,求f(2)+f(1/2 )+f(3)+f(1/3 )+……+f(2012)+f(1/2012 )的值
已知函数f(x)=x²/1+x² ,求f(2)+f(1/2 )+f(3)+f(1/3 )+……+f(2012)+f(1/2012 )的值
已知函数f(x)=x²/1+x² ,求f(2)+f(1/2 )+f(3)+f(1/3 )+……+f(2012)+f(1/2012 )的值
f(x)+f(1/x)=x²/1+x² +1/x+x²=1
所以
f(2)+f(1/2 )+f(3)+f(1/3 )+……+f(2012)+f(1/2012 )
=1+1+……+1
=2011
f(a)+f(1/a)=1.
结果就是2012-1=2011