答案是:a^2-1998a+1999/(a^2+1)=1998;解答过程如下:
因为a是方程的根,所以a^2-1999a+1=0
==》a^2+1=1999a
同时除以a ==》a+1/a=1999
对于a^2-1998a+1999/(a^2+1)=(a^2-1999a+1)+(a-1)+1999/(a^2+1)
=(a^2-1999a+1)+(a-1)+1999/1999a
=0+(a-1)+1/a=a+1/a-1=1999-1=1998
答案是:a^2-1998a+1999/(a^2+1)=1998;解答过程如下:
因为a是方程的根,所以a^2-1999a+1=0
==》a^2+1=1999a
同时除以a ==》a+1/a=1999
对于a^2-1998a+1999/(a^2+1)=(a^2-1999a+1)+(a-1)+1999/(a^2+1)
=(a^2-1999a+1)+(a-1)+1999/1999a
=0+(a-1)+1/a=a+1/a-1=1999-1=1998